Introduction to Complex Analysis
Introduction to Complex Analysis MATH 185
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Notes for Math 185 Conic Sections in the Complex zplane September 1 2006 345 pm Straight Lines The straight line through distinct points b and bc is the locus of points z satisfying Imz7blt 0 or equivalently Imz7b 0 wherein E is the complex conjugate of 9 The distance from other points z to f is lz7 l lImz7bEllcl because the point on f nearest 2 is z 7 1Imz7b z 7 19 Imz7blt Where do two straight lines intersect Unless they are parallel they intersect in just one point Let us nd it for lines whose equations are ImEz7b 0 and ImCZ7B 0 Put these equations into the form Ez7b 9Z7b and CZ7B QE7E and then eliminate E to get the intersection point z CImGb 7 cImCB ImQ Parabolas The Parabola whose directrix is f and whose focus is at f not on f is the locus of points z equidistant from f and from these points z satisfy the lfree equation lz7fl2 lz7 l2 zue iz7fi2 Imz7be2iei2 or H2 Imz7bE2iei2 At rst sight this equation factors into two equations 27 f l ilz7 l but no value of 2 can satisfy lz7fl 7lz7 l so long as f does not lie on f The parabola lies entirely on the same side of f with f since lz7fl gt lz7 l when 2 and f lie on opposite sides of Therefore when 2 lies on the parabola Imz7b and Imf7b must have the same nonzero sign and SO lzifl Sign1mfb39939ImZb399l9l We resume now the discussion of solutions to problems issued on 28 Aug 2006 They are Problem 6 Let distinct points i f be the two F oci of a Central Conic Section drawn in the complex zplane This curve is classi ed according to the equation satisfied by a real constant C and complex variable 2 on the locus Ellipse lz7fl lzfl ZQlfl 2 2m Hyperbola 72m S lz7fl 7 lzfl iZQlfl S 2m a Show how the inequalities descend from the Triangle Inequality lzwl S lzl wl b Show why 2 if it satis es one of the foregoing equations must satisfy also m2 m2 7 RezflflC2 7 CW 0 This equation is free ofthe square root implicit in c For which values Q do these conics degenerate to straight lines or line segments d What happens to the ellipse if f 7gt 0 and Q 7 00 while lel stays constant e Show that the hyperbola s Asymptotes equation is lzfl2 7 Rezf l2 0 Problem 7 A conformal map 2 lt7gt w is defined by two nonzero constants f and 5 thus 2 fwB Bw2 or equivalently wB zf i W zf2 7 l a Show that z i f fw i DZ215w and thus express the conformal map more symmetrically via the equation 2 fz 7 f w Bw 7 152 b Locate the map s Poles points in one plane mapped to co in the other Prof W Kahan Page 14 Notes for Math 185 Conic Sections in the Complex zplane September 1 2006 345 pm 0 The map is mostly 2tol in that it takes w W1 and w Bzwl to the same 2 Locate the map s Critical Points where the map is ltol instead of 2tol d Show that pairs of concentric circles in the wplane map to ellipses in the zplane thus For any real constant 2 gt 0 both circles lwl MK and lwl lBlQ map to the ellipse lzfl lzifl Q lQlfl except for a degenerate case in which the circle lwl 3 maps twice to the line segmentjoining foci if e Show that pairs of straight lines through the origin in the wplane map to hyperbolas with foci if in the zplane thus For any xed angle 0 the straight line through 0 in the wplane traced by w QBexpOQ as 2 runs through all real values maps onto the hyperbola lzf 7 lzifl i2lflcos0 replacing 0 by 70 yields another line mapped upon the same hyperbola except for two degenerate cases which of lines mapped to lines or line segments f At what angles do the foregoing ellipses and hyperbolas intersect Central Conics HighSchool Analytic Geometry in the Complex Plane Let i f be at the foci of a central conic section drawn in the complex zplane This curve is classi ed according to the equation satis ed by a real constant u and complex variable 2 on the locus Ellipse lzifl lzfl 2pm 2 2m so u 21 here Hyperbola 72m S lzifl 7 lzfl i2ulfl S 2m so 71S u S 1 here The inequalities come from the Triangle Inequality it i wl S ltl lwl for example try t zf and w 27f or t 2f etc The i sign in the hyperbola distinguishes its two branches On the hyperbola s asymptotes 2 satis es Rez f iulzfl or equivalently Imz eiWCCOSW f 0 where 12 fl For both conics the ends of the major axis are at iuf The ends of the ellipse s minor axis are at inuzil f To translate the center of the conic from 0 to b merely replace 2 by 27b in all equations above and add b to the foci and to the axes ends There are some degenerate and limiting cases Ellipse with f 0 lt um r is a circle ofradius r u 00 with u l is the line segment from f to if Hyperbola with f 0 is the whole plane with u 0 i f is the right bisector ofline segment f to if with iu fl is a halfline emanating from i f To rid the conic s equation of the square roots implied by we multiply together all the equation s four distinct conjugates namely ZHlfltlZifltlztfl 0 ZHlflilZiflilZ l 0 ZHlfl lzifl t lztfl 0 and ZHlfltlZiflilztfl 0 Prof W Kahan Page 24 Notes for Math 185 Conic Sections in the Complex zplane September 1 2006 345 pm only one of which can be satis ed in nondegenerate cases to get one equation that after division by 716lle simpli es to 02 m2 7 m 39Rezf2 e m2 0 This one equation is the equation of if HZ gt 1 then an ellipse if HZ 1 then the straight line through if twice if 0 lt HZ lt 1 then both branches of an hyperbola if u 0 then the straight line rightbisector of the line segment joining i f twice To get the equation of a circle of radius r gt 0 centered at 0 divide the equation by HZ and let f 7gt 0 and M7 00 while keeping ulfl r gt 0 The hyperbola s asymptotes equation is lz f l2 7 Rezfu2 0 which factors into the two equations lzfl iRezfu of two Vshaped figures consisting of halflines emanating from 0 making angles of iarccosu with the two line segments from 0 to if A Conformal Map of Lines onto Parabolas For any complex constant f a lt02 conformal map 2 lt7gtw is de ned either by z f w2 or equivalently by w The map is ltol at its two Critical Points 2 f w 0 and z oo w oo Every distinct pair in of parallel straight lines in the wplane maps to a parabola P twice in the zplane thus For any complex constants B and C i 0 the locus of w B Cu as u runs through all real values is a straight line H whose equation is Imw7BC 0 Parallel to H in the wplane is 7H the locus of w 7B CH whose equation is ImwBC 0 Both in are mapped to the zplane by z f w2 onto a parabola P whose equation is H Rezefycz 21mBC2 lCl2 f is the focus of P and its directrix is the line through f 7 2CImBC2 parallel to 1C2 whose equation is Re z C2 21mBC2 0 P degenerates into a halfline emanating from f when ImBC 0 A Conformal Map of Circles and Lines onto Central Conic Sections A conformal map 2 lt7gt w is de ned by two nonzero complex constants f and 5 thus 2 wB Bwf2 or z fz 7 f w Bw 7 152 0r wB zf 4 w zf2 7 1 Two points w and Bzw in the wplane map tofrom the same point z in the zplane except for the two Critical Points 2 if w iB where dzdw 0 Poles are at z 00 when w 0 or 00 but because the map s multiplicity does not change in the poles neighborhoods the poles are not critical points Prof W Kahan Page 34 Notes for Math 185 Conic Sections in the Complex zplane September 1 2006 345 pm Pairs of concentric circles centered at 0 in the wplane map to ellipses twice in the zplane with foci at i f as follows For any constant 2 gt 0 both circles lwl W9 and lwl lBlQ map onto the ellipse whose equation is lzfl lz7fl Q lQlfl The circle with Q 1 maps twice to the line segment joining the foci i f The two annuli between the outer circles and the inner one map to the interior of the ellipse traced twice To see why this is so observe that in general 2 i f fwB i 2 Bw2 fw i DZ215w so when w is on the circle lwl lBlQ for any constant 2 gt 0 w H elf62H w 159 w7B22 7 Elf52H 2le 2W2 2 elf62H lBl292 W s2 mm as claimed Replacing Q by 19 in the wplane changes nothing in the zplane The circle lwl lBlQ can be generated by setting w lBlQexpOQ and letting angle 0 run through all real values between 77E and HE An alternative that requires no transcendental function is to set w lBlQu 1u 7 1 and let u cot02 run through all real values Pairs of straight lines through the origin in the wplane map to hyperbolas twice in the zplane with foci at i f as follows For any xed angle 0 a straight line through 0 in the w7plane is traced by w QB exp10 as 2 runs through all real values here exp10 0050 1sin0 stays constant on the line Two lines are mapped onto each hyperbola lzfl 7 lz7fl iZlfl cos0 To see why recall that in general 2 i f fwB i 2 Bw2 fw i DZ215w so on the line w QBexpOQ we find that w 7 lzefl lfl39 was 7 MW W wl 2m ReBwlel 2m signQcos0 Replacing 0 by generates the same hyperbola in the zplane from two lines in the w plane that are distinct except in the degenerate cases 0 0 or in and 0 int2 Because the straight lines through 0 intersect circles centered at 0 orthogonally in the w plane and because the conformal map preserves angles of intersection except at its critical points the hyperbolas and ellipses described above intersect orthogonally in the zplane too Where do two different conics intersect They can intersect in at most four points Finding these generally entails solving a quartic or cubic equation that can t be solved using only finitely many real operations each restricted to one of 7 IL But special cases exist like Centers of the Circles of Apollonius of Perga ca 200 BC A story for another day Prof W Kahan Page 44 File ConcyclicS Solutions for Math H185 February 8 2008 841 am When are three distinct points C ollinear When are four distinct points C oncyclic Here collinear means on the same real straight line and concyclic means on the same circle including maybe a straight line regarded as a circle with in nite radius The following problems elaborate problems 26 and 27 at the end of 12 of Basic Complex Analysis 3rd ed by JE Marsden amp MJ Hoffman 1999 0 Solve the equation w 7 tz 7 t z 7 tz 7 w for one of the three complex numbers t w and z in terms of the others to show that as points in the complex plane they are situated at the VeI ticeS of an equilateral triangle Note that the equation is invariant under shifts of origin Solution Shifting the origin to t by subtracting t from every variable amounts to setting 2 t 0 in the equation turning it into wz zz 7w whence z 7wz w2 0 Therefore z qw for q 1 i 1V3 eim3 choose either value Multiplication by q performs a rotation through TE3 60 w 7z qw is rotated the same amount in the opposite direction And since lql lql 1 we find that lw 7zl lwl lzl so the points z w and t 0 are the vertices of an equilateral triangle as claimed 1 Given three distinct vectors t w and z interpreted as displacements from some origin 0 to three distinct points in a real linear space of arbitrary dimension what condition do these vectors satisfy just when the points are collinear Why Solution The three points are collinear just when w 7t and z 7t are Linearly Dependent either parallel or antiparallel in this case which means w 7t z 7t for some real scalar LL 0 Replacing the equation s differences by any other two distinct differences among the three vectors merely changes the scalar If the three vectors are actually complex numbers then the equation says that any difference quotient like w 7tz 7t is real This one is negative just when t lies between w and z on the line 2 Given only the three Euclidean distances Ht7wH Hw7zH and Hz7tH between those three distinct points what condition do the distances satisfy just when the points are collinear Why Solution The distances satisfy the Triangle Inequality it says that the biggest of the three must equal or exceed the sum of the other two Equality occurs if and only if the points are collinear and appropriately ordered Hz7tH Hw7zH Ht7wH just when w lies between t and z on the line Were distance not Euclidean equality would be necessary but not necessarily sufficient for collinearity See ltwwwcsberkeleyedukaahanMathHl10NORMlitepdfgt Prof W Kahan Page 14 File ConcyclicS Solutions for Math H185 February 8 2008 841 am 3 Given four distinct vectors q t w and z interpreted as displacements from some origin 0 to four distinct points in Euclidean 3space what conditions do these vectors satisfy just when the points are concyclic Why ltwwwcsberkeleyedukaahanMathHl lOCrosspdfgt may help Solution Temporarily let s assume no three of the four given points are collinear and simplify the algebra by translating the origin 0 to q say by subtracting q from every given vector Through any three noncollinear points p7v p and pu in Euclidean 3space goes exactly one circle and its center is known according to problem 8 in 9 of Crosspdf to lie at c 7 p HvH239uuT 7 HuHZ39WT39V uHvH239HuH2 7 vTu2 From any four distinct concyclic points we may choose two subsets of three arbitrarily and each subset must lie on the same circle and thus determine the same center via this formula Let our two subsets have say p q0 ut7qt Vq7w7w and p 0 ut Vq7z7z Then substitution into the formula above provides an equation Hzllz39lltll27th239HWH239ttT7 HtHZ39WWT39t7W 7 HWHZHtll27th2HzH2ttT7 HtHZ39zzT39t7z that simpli es drastically I hope Alternate approach Given three noncollinear points p7v p and pu in Euclidean 3space their circle s center 2 2 2 c 7 p V 39u 39V39HuH uHvH Hv uH Here V u rewrites qu to turn crossproducts into associative matrix multiplications with Jacobi s Identity v u V u 7u V Grassmann s Identity v u uVT7vuT Substituting our subsets into the equation for c produces an equation 2 2 2 2 2 2 w t wHtH 7 tHwH Hw tH 7 z t zHtH 7 tHzH Hz tH I t that Jacobi s Identity turns 1n 0 w t wHtH2Hw tH2 7 z 39t 39z39HtH2Hz 39tll2 7 t z tHzH2Hz tH2 7 t w tHwH2Hw tH2 4 Given four distinct complex numbers q t w and z regarded as points in the plane what condition do these numbers satisfy just when the points are concyclic Why Solution A condition both necessary and sufficient that the four given points he on the same circle or straight line is that w7zq7tq7zw7t be a nonzero real number Its sign depends upon the order of the points Why and how will be explained next First the condition will be proved necessary Suppose the given points he on a circle or on a straight line regarded as closed by one point at 00 Two configurations are possible The rst has q t w and z in consecutive order around the circle they are successive vertices of a concyclic quadrilateral whose opposite angles must be Supplementary as shown rst below Prof W Kahan Page 24 File ConcyclicS Solutions for Math H185 February 8 2008 841 am Convex and Crossed Concyclic Quadrilaterals In this case wizq7z OLele for some 0L gt 0 and 0 S 9 S T and qitw7t Be39m e for some 5 gt 0 Multiply to con rm that wizq7tq7zw7t OLBe 17E 70LB lt 0 In the second con guration q and t separate w and z as shown second above Swapping z and w changes the picture but not much of the proof Now wizq7z OLele and qitw7t Be le so that wizq7tq7zw7t OLB gt 0 confirming the condition To prove the condition suf cient reverse the argument and depending upon the sign of the real quotient exploit either the fact that a convex quadrilateral whose opposite angles are supplementary must be concyclic or else the fact that two triangles with a common base on the same side of their other vertices where the angles are equal must have the same circumcircle 5 Given only the six Euclidean distances HqitH Hqin HqizH Htin HwizH and HzitH between four distinct points in Euclidean nspace what condition do these distances satisfy just when the points are concyclic Why Looking up Ptalemy s Inequality may help Solution The condition is that the largest of the three products HwizH Ht7qH HzitH Hw41H and HwitllHZAIH equal the sum of the other two It comes from Ptolemy s Inequality which says that HwizH Mtqu HzitH Hw41H 2 HwitllHZAIH with equality just when the four points he on the same circle or straight line and are so ordered that q and z separate t and w A mnemonic for this inequality treats the four points as vertices of a tetrahedron on any closed path that traverses some four of the tetrahedron s six edges The path is a quadrilateral the sum of the products of its opposite edgelengths exceeds the product of the two untraversed edge lengths except when the tetrahedron has collapsed onto a concyclic quadrilateral or straight line segment in a 2dimensional plane in the ndimensional space t Ptolemy s Inequality llwrzll39ll Hlll Hzitll39llwiqll 2 llwrtll39llz ll with equality just when the tetrahedron s vertices lie on the same circle or straight line so ordered that q and z separate t and w Prof W Kahan Page 34 File ConcyclicS Solutions for Math H185 February 8 2008 841 am Problem 5 s solution is a byproduct of a deduction of Ptolemy s inequality from the Triangle Inequality applied to Inversion in the Unit Sphere Such an inversion is an invertible relation y xHxH2 or equivalently x yHyH2 wherein the Euclidean norm Hull Vu39u Either equation between x and y implies the other except perhaps when a vector is 0 or 00 Moreover the identity xHxH2 7yHyH2 E H x 7y H HxHHyH for arbitrary nonzero x and y is a quick consequence of the Euclidean norm s de nition No generality is lost by a shift of origin 0 to q say that subtracts q from each of the four given vectors Doing so turns Ptolemy s inequality into the equivalent inequality erzll39lltll t HZ tH39HWH 2 ertll39llzll Its con rmation begins with the Triangle inequality which asserts that llwlle27tHtllzll s IIwiiwii27ziizii2II llzHzll27tHtllzll Into this substitute the identity above and multiply by HtHHwHHzH to nish the con rmation The Triangle inequality above and therefore Ptolemy s becomes an equality just when wHwH2 7 zHzH2 M zHzH2 7 tHtHZ for some M gt 0 this is equivalent to the equation zHzH2 wHwH2 M39tlltll21 H Think of M as avariable parameter As it increases from 0 to 00 it puts zHzH2 in a straight line segment running from wHwH2 to tHtHZ Inverting the inversion puts z into an arc that runs from w to t That arc is part of the image by inversion of a straight line The image is a circle through 0 or if the line passes through 0 the line itself with 0 and 00 swapped This assertion can be con rmed by restricting attention to a plane 2dimensional subspace of the n space containing 0 and the line Choose orthonormal X ycoordinates in which the line s equation is say y T a constant and observe that inversion s image X y EN T lE2 112 runs along a locus whose equation deduced by eliminating the free parameter E turns out to be 11012 y2 7 y 0 the equation ofa circle through 0 0 or if T 0 the same straight line with 0 0 and 00 0 swapped After restoring the origin we infer that equality in Ptolemy s inequality is a condition necessary for four given distinct points q t w and z to lie on the same circle or straight line in such an order that q and z separate or are separated by t and w But problem 5 provides only the siX distances between the points not their order Without order information the condition necessary for our four points to be concyclic is that the largest of the three products erzll39llHIH Hzrtll 39HWAIH and ertll 39llz ll equal the sum of the other two This condition is also suf cient because the reasoning above is reversible Prof W Kahan Page 44 NOTATION IN MATH 185 D ave Pen neys Here is some notation we have seen thus far c R Q Z N XCY fX Y zHy X XgtltY gt lt means the set of complex numbers means the set of real numbers means the set of rational numbers means the set of integers means the set of natural numbers means that the set X is a subset of Y This means the set X is contained in the set Y For example7 N C Z C Q C R C C means f is a function from the set X to the set Y means the element x maps to the element y means that z is an element of X is the cartesian product of X and Y It is the set of all ordered pairs my such that z E X and y E Y XgtltYyz X andyEY me ans for all 77 means there exists means such tha means unique or factorial if after a number means implies is the Boolean symbol for not is the Boolean symbol for and is the Boolean symbol for or File DSpr2p43 Solution for Ex 2 p43 of Sarason s Notes October 16 2006 1018 am Exercise 2 p 43 Let a be a complex number of unit modulus and c an irrational real number Prove that the values of a9 form a dense subset of the unit circle Solution Say a exp1tt for some real LL then the values of a9 in question are the values of expou 2nTEc as n runs through all integers These are the same values as are taken by expoLN 2nc 7 kTE as k runs through integers like nc the biggest integer no bigger than nc Let fx x7 x denote the fractionalpart ofa real number x so that 0 S fx lt l and fxy ffx fy To solve the problem we must explain why the values taken by fnc as n runs through all integers are dense in the interval 0 lt fnc lt l The proof goes by contradiction Since every member of the sequence n9n 1 23 lies strictly between 0 and 1 else 9 would be rational nothing is lost by rst joining the ends of the interval 0 l to wrap it around a circle and then treating the sequence s members as points on that circle Suppose for argument s sake that the sequence were not dense Then some open arc A on the circle would be empty contain no points of the sequence Let A be the widest empty arc containing A such a maximal arc A would have to exist because its endpoints would be limits of monotonic sequences of endpoints of ever wider empty arcs For each m 0 l 2 3 the arc fA 7mc all points f0c7mc generated by letting 0L run through A would have to be empty too since fnc could fall in fA7mc only if fnmc fell in fA A All the arcs fA 7 mg would be di erent because the equation foc7 mc foc 7 kg iftrue for di erent integers m and k would imply that c is rational All the arcs fA 7 mg would be disjoint nonoverlapping too because the overlap of fA 7mc with fA 7 kg for m gt k would imply the overlap of fA 7 m7kc with fA A contradicting the maximality of A But no circle can hold in nitely many disjoint open arcs fA 7 mg of equal nonzero widths so they cannot be empty open arcs after all the sequence fncn 133 is dense as claimed I believe this proof can be traced back to L Kronecker in the second half of the 19th century Satvik Beri s Solution First this problem will be reduced to a search for integers m and n that make l r 9n 7 ml arbitrarily small after irrational c and any arbitrary real r have been xed After this the two sought integers will be proved to exist What values are taken by a9 Since tat l we can write a e H T 2n for some fixed real LL between in and any integer n and then by definition the set of values of 519 consists of the values taken by exp1cu 2nTE as n runs through all integers Since all of these values have magnitude 1 they all lie on the unit circle Any point on the unit circle can be expressed as e13 for some real 5 between in inclusive To prove that the values of exp1cu 2nTE are dense in the unit circle we have to show that given any such 5 there are integers n that make differences l exp1cu 2nTE 7 e18 l arbitrarily small This can be done because exp1 is a continuous periodic function of a real argument by finding integers n that make l cu 2n 3971 7 5 mod 27E l arbitrarily small which is tantamount to searching after 9 and LL have been fixed for integers n and m that make l cu 2nTE 7 B 7 2m 397E l arbitrarily small for each given real 5 Prof W Kahan Math 185 Page 12 File DSpr2p43 Solution for EX 2 p43 of Sarason s Notes October 16 2006 1018 am Divide out 27c and set r 9H 7 B27E now we search for integers n and m that make l r 9n 7m l arbitrarily small after real r and irrational 9 have been fixed What arbitrarily small means is that having chosen any big integer K gtgt 2 we can nd integers m and n that make lr cniml lt l2K What follows will prove that such integers m and n exist Let fX be the fractional part of any real X as described above and break the open interval 0 lt X ltl into K separated fragments 0 lt X lt lK lK lt X lt 2K 2K lt X lt 3K and l 7 lK lt X lt 1 As k runs through 1 2 3 K the K values fkc scatter into the insides of those fragments because no value fkc can be rational when 9 is irrational At least one value f ltc must fall into the first fragment 0 lt f ltc lt lK otherwise some two of the K values fkc say fk1c and fk2c for 0 lt k1 lt k2 S K would have to fall into the same one of the other K71 fragments making l fk2c7 fk1cl lt lK and implying either 0 lt fk2399fk1399 f1lt21lt139ltlt UK or else 0 lt fk1399fk2399 f1lt11lt239ltlt 1K Either way 0 lt k k27k1lt K or 0 gt k k17k2 gt 7K has 0 lt fkc lt lK as claimed Having chosen this k let j be the integer positive or negative nearest irfkc so that lj rfkcl S 12 Then set integers n jk and m jkc jkc ijfkc to get if t 93911 ml l r tj39k39 j 39k399 j39f1lt39lt l f1lt39939ljtrf1lt399lS f1lt3992 lt12K End Prof W Kahan Math 185 Page 22 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am The Players Real variables are r s u v x y 0 u 5 p Complex variables are t r 1s w u 1v z x 1y where 1 Wil Real functions are g h P Q X Y Complex functions are f g h p q usually Differentiable Functions We suppose that complex fz fx 1y gx y lhx y and that g and h are real continuously differentiable functions of x y in the sense that dgX y glOXa y39dX gorX y39dy and dhX y h10X y39dX horX y39dy with continuous partial derivatives g10x y Bgx yBx h01x y Bhx yBy Here the differentials d are to be determined by the Chain Rule as follows Substitute arbitrary differentiable real functions X05 and YB of some real variable 5 for x and y respectively then every d can validly be replaced by d dB in the equations above for dg and dh Is f a differentiable function of z The answer would already be Yes if instead of complex variable and function z and f were vectors in 2dimensional real Euclidean space say rowvectors z x y and f g h Then df dg dh and dz dx dy would have to satisfy dfz dzf39z for a 2by2 Jacobian matrix f z whose elements were the four partial derivatives of gx y and hx y above f39 gm 0 But this f z could not lie in gm 1 the same space with z and fz to write dfz f zdz would be wrong because no 2 by2 matrix f z can premultiply arowvector dz Multiplication of complex variables is commutative wz zw If f39 is to be a complex function whose multiplication by other complex functions and variables is commutative too the matrix f cannot be an arbitrary matrix We have already seen that special 2by2 matrices Wl la EM hl ivu yX hg are algebraically isomorphic with complex variables and functions w u 1v z x 1y f g 1h respectively in so far as rational algebraic operations are concerned That these matrices commute only with their own kind is easy to confirm so matrix f has to be a special 2by2 matrix too h01 glo and g01 ihlo The last two equations are the famous Cauchy Riemann Equations about which we have just deduced If the complex function fz of the complex variable 2 has a complexvalued derivative f39z satisfying dfz f39zdz for all complex dz then fx 1y s real and imaginary constituents gx y Refx 1y and hx y Imfx 1y must satisfy the Cauchy Riemann Equations Bhx yBy Bgx yBx and Bgx yBy 49hx yBx and then f 39x 1y Bgx yBx 13hx yBx Bhx yBy 7 13gx yEy Example The function le Zz has no complex derivative at z 0 because then the real and imaginary parts of ix lyl2 x2 y2 10 violate the CauchyRiemann Equations lzl2 is a differentiable function of real variables x and y since dlz2 Zdz zdZ 2xdx 2ydy Prof W Kahan Page 1 Notes for Math 185 Derivatives in the Complex 2plane May 16 2008 413 am Exercise 1 At every complex 2 0 the function l2 2l2l2 has a complex derivative 7122 so the real and imaginary parts of lx 1y xx2 y2 17yx2 y2 must satisfy the CauchyRiemann Equations and they do check them out Exercise 2 Suppose f2 is differentiable over an open domain in the 2plane mapped by the assignment w f2 to some region in the wplane Show that the images fC and fc in the wplane of any two smooth curves C and c in the 2plane have the same angles of intersection except if C and c intersect at a critical point 2 where f392 0 This is why such a map is called a Conformal map Hint What do orthogonal 2by2 matrices do How does a given expression for f2 get turned into an expression for f 392 If f2 is an algebraic function the rules for symbolic differentiation turn out to be the same for complex as for real expressions The rst rule worth knowing is that the derivative f392 is the limit as w gt 2 ofthe Divided Difference fl2 w f2 7 fw 27w simplified symbolically which as we shall see simplifies symbolically to a polynomial in 2 and w if f2 is a polynomial in 2 a rational function of 2 and w if f2 is a rational function of 2 or an algebraic function of 2 and w if f2 is a continuous algebraic function of 2 after the division by 27w has been carried out After that simplification f392 fl2 2 Note fl2 w is a function ofunordered pair 2 w so that fl2 w flw 2 There is no standard notation for fl2 w other authors use 2 wf or f2 w or Af2 w or Example For any integer N 2 0 the divided difference of f2 2N simplifies to fl2 w 21 sk 3N2 1 wN k after which f392 fl2 2 N39ZNTI the same for complex as for real 2 The same for complex as for real generalizes to arbitrary rational functions f2 because as is easily verified if f2 p2 i q2 then fl2 w pl2 w i ql2 w respectively and if fz p2q2 then me w plan wq2 pwqlez w and if fz p2q2 then me w plan wqw e pwqlez w q2qw The latter two formulas have alternatives generated by the identity fl2 w flw 2 The second rule worth knowing is the Chain Rule If pw and q2 are complex differentiable functions of complex arguments then f2 pq2 has a complex derivative f392 p39q2q392 This follows directly from the Chain Rule for differentiable vectorvalued functions of vector arguments rst treat 2 q p and f as 2vectors and then convert derivatives from special 2 by2 matrices back to their complex form Another way to go is the Divided Differences ChamRule If fzpqz then flzwplqzqwqlzw Prof W Kahan Page 2 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Example If fz 32 122 then ffz w 32lz 3wlw3 7 lzw and then f z23z lz3 7122 since pt t2 has plst7st Analogous to the partial derivatives Bpz tBz and Bpz tBt of a differentiable function pz t are its partial divided differences pizw t pzt 7 pwt z7w simpli ed symbolically to eliminate z7w and similarly piz wt pzw 7 pzt w7t Once again if pz t is a polynomial in z and t so are its partial derivatives and divided differences Likewise if p is arational function To deal with algebraic functions we must introduce Implicit Divided Di erencing If fz is a root of the equation pz fz 0 then flzw 7plzw fzplw fzfw This follows by elementary algebra from pz fz pw fw and the de nitions of mi Then it is tempting to assume f is continuous let fw gt fz as w gt z and deduce that f 39z 49pz tBz Bpz tBt evaluated at t fz Algebraic functions f aren t quite so simple as we ll see after we discuss Implicit Functions Example When pz t t2 7 z a particular root of pz fz 0 is fz V2 and this yields ffz w lVE VW Since V2 is continuous except as z crosses a slit along the negative real axis in the zplane dVZdz f39z limwaZ fiz w fiz 2 lZVE except on the slit Continuity is essential here had we taken fz V2 and fw 7Vw as roots of pz fz 0 and pw fw 0 respectively we would have obtained a divided difference fz 7 fw z7w lVZ 7 VW with 00 instead of the correct derivative for its limit We mustn t make this mistake when we discuss Implicit Functions Just as the derivative is the limit when it exists of a divided difference the divided difference is an average of the derivative when it is exists and is continuous thus Hermite s formulation ffz w fol f39w z7wudu In other words ffz w is the uniformly weighted average of f 39 on the line segment joining z and w Hermite s formulation needs no division by z7w so it works also for perhaps vectorvalued functions fz of vector arguments z for which we find Exercise 3 Deduce from Hermite s formulation that fz 7 KW mz wz7w flz w aw z and flz z 7 f z Then show that the divided difference of a polynomial function of a vector variable s elements is also a polynomial function of the vector s elements but Hermite s divided difference of a rational function of a vector variable s elements can be a nonrational function of them alas This is why derivatives drove divided differences out of fashion fI has more variables than f has and Hermite s fI can be transcendental when f is still rational or algebraic Other ways to define ff exist that remain rational or algebraic respectively and satisfy Exercise 3 s equations but they depend upon the coordinate system in the space of f s argument and lack properties that will be revealed in the following digression Prof W Kahan Page 3 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Digression Interpolation with HigherOrder Divided Differences Many Numerical Analysis texts and all texts about Finite Di quoterences devote at least a chapter to divided differences of higher than rst order Here they will be surveyed only enough to convey a sense of their utility Proofs will be omitted during this digression Alas no fully satisfactory notation exists for higherorder divided differences so the literature does not agree upon one For n 0 l 2 3 let us write fnx for the nth derivative of fx at argument x and write flnlx0 x1 xn for the nth divided difference of fx over argument nltuple x0 x1 xn Yes f0 floT f f1f39 and fl1T fl Here for n gt 0 is Hermite s formulation nlflnl xo x1 xn is the uniformly weighted average of fnx as x ranges overa simplex whose n1 vertices are x0 x1 xn n 1 Cir 12 Clair flnloxo x1 x 010 0 IO f xo Zqjltxjinomqnmdqsdqqul 391 The simplex degenerates collapses if the vectors xi 7 x0 for j l 2 n are linearly dependent but the foregoing integral remains valid and most important independent of the order of the xj s For example when n 2 the simplex is atriangle and as you should verify by manipulating the integral Wax y 2 Wax y 2 Way 2 x Way x 2 and so on And fllx x x fquotx2 just as flnlx x x fnxn in general Divided differences supply remainders to Taylor s formula if z xh then fz fx f xh f xh22 f3xh33 fn 1xhn 1nil ftnqu x x zhn in which the nltuple x x x 2 has x repeated n times followed by z This formula is valid for vector arguments x and 2 provided hk is interpreted to mean k repetitions of the vector h zix as the arguments for the klinear operators fk and flkl which act linearly on each of k vector arguments thus f kx v1 v2 vk is a linear function of each vector vJ separately and because of the Clairault Schwartz theorem that says the order in which multiple differentiations are performed doesn t matter if derivatives are continuous the order of vectors vj after fkx doesn t matter either Divided differences allow Taylor s formula to be generalized to Newton 3 Divided Di erence formula of which the following instance is simplified to dispense with subscripts fz fY flXy39zeY fl2lwxy39zeX39zey fl3lvwxy39zew39zex39zeY fl4lvwxyz39zeV39zW39zex39zeY If the last remainder term in this formula and in Taylor s is small enough as it is if f4 is small enough or if z is close enough to v w x and y then f may well be approximated by the polynomial function of its argument consisting of the previous terms they constitute an Interpolating polynomial in 2 which matches the values fz takes when 2 y z x z w and z v If some of these arguments v w x y coincide certain of the polynomial s derivatives match f s at those repeated arguments Note Newton s and Taylor s formulas work for nonscalar vector arguments v w x y 2 only with Hermite s formulation Genocchi used it too of divided differences and then the polynomial s degree need not be minimal Prof W Kahan Page 4 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am n flnlx0 x1 xn is an average which must take a value inside the convex hull of the values taken by fnx as x ranges over the convex hull of the points x0 x1 and xn but that average value need not be a value taken by f n anywhere When f and its arguments v w x y z are restricted to scalars real or complex much more can be said In the real case if f n is continuous it must take its average value at some place E inside any interval containing all of x0 x1 xn and then flnlx0 x1 xn fnEn This form for the remainder occurs frequently in the literature In general any expression for f n that yields bounds for its values over some domain provides a bound for the difference between f and an interpolating polynomial of degree less than n over that domain When f is a complex analytic function of a complex scalar variable though f may be real if its argument is real flnl can be bounded without computing fn provided if can be bounded over a suitable closed contour C in the complex plane C must enclose no singularity of f and yet enclose all the points x0 x1 and xn Let 52 Hon zixj then it turns out that Zlnflnl xo x1 xn C fzdzBz The contour integral implies that if every zixJi gt u while 2 runs on C whereon ifzi lt M then 2niflnlx0 x1 xnl lt Mlength c yum We prefer that this bound be small in order that an interpolating polynomial of degree less than n approximate f well but the bound cannot be made arbitrarily small because Mlength C generally grows ultimately rather faster than un as u increases so some skill is needed to choose C well enough to produce a bound about as small as possible The contour integral produces an explicit formula for flnl valid even if f is not analytic in the simple case that all the points x0 x1 and xn are distinct the formula is simply flnlx0 x1 xn ZJ fxJB39xJ in which B39xj is the product of all n differences xi 7 xk with k j In other words flnlx0 x1 xn can be computed from the values f takes at distinct scalar arguments without any recourse to derivatives However the foregoing sum is rarely a satisfactory way to compute flnl numerically Usually better is flnlaxo Xi xi H Wixo Xi xii iflnrllGXi x2 xnxoegtltn especially when points x0 x1 and xn are in monotonic order exploiting the observation that highorder divided differences are divided differences of lowerorder divided differences That recurrence generalizes to produce a Con uent Divided Difference flnlx0 x1 xn when some of its arguments coincide see Commun Assoc Comp Mach 6 1963 pp 1645 Exercise 4 This example shows Why Newton s formula used often to interpolate functions of scalar arguments is almost never used with vector arguments In the x yplane let rectangle R be given with diagonally opposite vertices at 0 0 and X Y What is the expected degree of the polynomial that Newton s formula would produce to interpolate fx y at the four vertices of R 7 The same interpolation is accomplished by the Bilinear polynomial px y f0 0 xflox 0 yfto 0Y xyfTTox 0Y If x y lies in R inside which f has continuous second derivatives show that fx y 7 px y xx7XBzfBx2 yinBzfBy2 for derivatives each evaluated somewhere in R Prof W Kahan Page 5 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Implicit Functions It is not obvious that every nontrivial polynomial equation pz t 0 must have roots t much less that they can be chosen in a way that makes them continuous functions of z The existence of as many roots tz as the degree in t of pz t is a famous theorem of CF Gauss that we shall prove easily later Their continuity is a theorem infamous for proofs that hide the difficulty of identifying correctly each of a number of singlevalued functions tz whose values may coalesce at critical points z where the equation has multiple roots For example what are the roots t of t3 7 z 0 if z 0 The Principal Cube Root often 213 is the root t with 77173 lt argt S TE3 if z reIQ with r 2 0 and 77E lt 0 S 7 then this 213 3Vre39Q3 It is discontinuous across the zplane s negative real axis 0 in whereon it takes nonreal values different from the Near Real Cube Root best denoted by 3V2 which is the root t with 7lW3 lt ImtRet S lW3 On all ofthe real axis this 3V is real and continuous with sig113x signx Elsewhere 3V2 is discontinuous only across the imaginary axis The two definitions of cube root disagree in the left halfplane there 3V2 77z13 Which definition is better Neither Consider the next example written simply What are the roots t of t3 321 7 22 0 if z 0 A widely used book tenders the formula t 213 1 Wmf l 7 Wmf without mentioning that its three cube roots must be chosen in a correlated way Principal Cube Roots for all three are never correct choices Exercise 5 Explain why NearReal Cube Roots for all three choices are better but imperfect if the book s formula read tz 7 3xE 3x1 Vm 3x1 7 Vm it would be wrong when 7Imz22 s Rez lt 0 i e between the imaginary axis and a parabola in the left half of the zplane This formula s three cube roots are chosen correctly only when the one chosen for z is the negative of the other two s product otherwise this formula s tz dissatisfies the equation t3 32 t 7 22 0 Exercise 6 Explain why Show for z 0 that a correct formula and perhaps the simplest is tz q 7 zq for each choice q of one ofthe three cube roots of l Wm 2 Show that replacing Wm by 7Vm merely permutes the three roots tz As 2 gt 00 one of the three roots tz approaches a finite limit what is it But no choice of cube root for q makes this root tz continuous over the whole zplane If q is the Principal Cube Root t is discontinuous as z crosses either the line segment 71 lt z lt 0 or an approximately hyperbolic curve in the left halfplane on which 2 x1y satisfies the equation 3x2 7 y22 78xx2 y2 If q is the NearReal Cube Root t is discontinuous across another approximately hyperbolic curve in the right halfplane this curve s equation is 2 2 2 4 4 My 73X Xy 7X In short no formula for a root tz of the equation t3 321 7 22 0 can be continuous on all the zplane every formula jumps as z crosses some curve joining the point 0 to at least one of 7l and co These three points are the equation s critical points where it has multiple roots triple at z 0 double at z 7l and co As 2 moves around a critical point the formula for a root would have to become multivalued if it stayed continuous To see why Prof W Kahan Page 6 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Exercise 7 Use the formula in Exercise 6 to trace the behavior of all three roots tz as z traverses a tiny circle around a critical point and watch two or three of the roots swap places as if playing Musical Chairs And yet wherever a root tz of the equation is Simple ie nonmultiple it can be obtained from a formula that is continuous in some open region perhaps small in the zplane This is a harbinger of what happens generally to roots tz of an analytic equation pz t 0 The simple roots local continuity is a consequence of a general Implicit Function Theorem If pzO to 0 and if pz t is continuously differentiable in a neighborhood of z Z0 and t tO and if Bpzo tBt 0 at t tO so tO is a simple root of the equation pzO to 0 then throughout some maybe smaller neighborhood of z zO a continuously differentiable function fz exists satisfying pz fz 0 and fzO tO and the derivative of f is f39z 78pz tBz Bpz tBt evaluated at t fz Proof You may have seen a theorem like this already for vectorvalued functions of vector arguments though then the place of Bpz tEt was taken by Bpz tBt 1 for a nonsingular instead of nonzero derivative The following proof for complex variables is similar The easy part of the proof is the formula for f39z which will follow from ImplicitDi erentiation 0 dpz fzdz Bpz fzEz Bpz tBttZf39z which will follow in turn from the formula for Implicit Divided Differencing above The hard part is proving that a continuous fz exists in some neighborhood of z zO To simplify the proof s notation let qt pz t for some fixed z so close to Z0 that lqtl is very tiny for all t close enough to tO fZO later we shall choose so close and close enough to ensure that the equation qt 0 has just one root t fz that close to fzO The needed closeness depends upon how wildly q39t Bpz tEt varies since it must be prevented from varying too much compared with q39tO which must be nonzero if z is close enough to Z0 because then q39tO is very near the given nonzero Bpz0 tEt at t tO These requirements come together in an important Lemma 1 If a positive constant u can be found to satisfy u gt lqtOqlt0tl whenever ltitol S u then at least one root t of the equation qt 0 also satisfies Vital lt u No more than one such root t exists if qlw t i 0 too whenever both ltitol lt u and lwitol lt u Proof When two such roots exist say qt qw 0 but w i t then qlw t 0 which contradicts the second hypothesis Therefore the lemma s nontrivial part is the inference from the first hypothesis that at least one root t exists To this end define Qt t 7 qtqlt0t Since the divisor cannot vanish Qt is continuous throughout the closed disk ltitol S u And Qt7tO 7qtOqlt0t so lQt7tOl lt u throughout the disk in other words Q is a continuous map of the closed disk into itself By Brouwer s Fixed Point Theorem Q must have a fixedpoint t Qt in that disk this is the root t we seek End of Lemma l s proof Prof W Kahan Page 7 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Readers who are still uncomfortable with the divided difference ql can resort to a very similar lemma stated exclusively in terms of the derivative q39 instead though its proof is longer Lemma 2 If a positive constant u can be found to satisfy u gt lqtOq39tl whenever t lies in the disk lt7t0l lt u then at least one root t of the equation qt 0 also lies inside that disk No more than one such root exists if also 1 7 q39tq39t0l lt 1 whenever t is inside that disk Proof We shall construct a trajectory t Ts by solving an initial value problem TO t0 and dTds qTqT for 0 S s S S This differential equation must have at least one complexvector solution Ts for all real s in some interval of suf ciently small positive width S because qtqt is a continuous function so long as t stays inside the disk wherein q39t 0 see Peano 3 Existence Theorem in a textbook about Ordinary Dijfkrential Equations And so long as TS is inside the disk S can grow a little How big can the width S grow and still have the trajectory graph of T strictly inside the disk lT7tOl lt u We are about to nd that S can grow arbitrarily big First let S be any positive value for which the trajectory Ts stays inside the disk throughout 0 S s lt S Along the trajectory we nd dqTsds q39TsT39s qTs whence follows that qTs eqto Then the length of the trajectory from s 0 to s S must be If ldTsdsl ds If lqTsq39Tsl ds If moomo eds lt If H39e 39ds lt H Evidently TS lies strictly inside the disk for every positive S Let S gt 00 to nd Li ldTsdsl ds Li iqtOq39Ts i ends lt Li H39e 39ds H Therefore Too lies strictly inside the disk too And qToo 0 In other words the trajectory t Ts ends strictly inside the disk at a root t Too of the equation qt 0 Now we know the disk contains at least one root can it contain two if also 1 7 q39tq39t0l lt 1 whenever t is inside that disk No in fact now qt can take no value zero or not more than once inside that disk Otherwise we could nd T inside that disk too with qT qt but T i t in which case we would also nd 1 l1 qlTa tq39tol l U 1 1390 t TitSq39todS l S U l1 1390 t T4SQ39tolds lt 1 This contradiction ends the proof of Lemma 2 Lemmas l and 2 provide ways to con rm that a root t of qt 0 lies within a distance u of tO but no way to nd tO nor u One way to seek u given q and t0 is to choose an M somewhat bigger than lqtOq39t0l and nd an overestimate u of max t7t0 SMlqtOq39tl or of max HO SM lqtOqlt0tl hoping that this u S M But it need not be so Exercise 8 For qt t2 7 l and t0 4 show that no u can be found to work in Lemma 2 but Lemma 1 works if 3 lt u lt 4 For qt exp1t l and t0 313 show that u 27E in Lemma 2 establishes the existence of a root t satisfying lt7t0l lt u but not its uniqueness Show that the second hypothesis of Lemma 2 implies the second of Lemma 1 and if qt is a real function of a real argument t and if tO is real and the disk is an interval in which a real root t may lie then show the first hypothesis of Lemma 2 implies the rst of Lemma 1 Prof W Kahan Page 8 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Back to the proof of the Implicit Function Theorem Given that pzO to 0 and for any given suf ciently tiny tolerance u gt 0 we seek a tolerance B so small that for every 2 in the disk lz 7 20 lt B the equation pz t 0 has just one root t in the disk it 7tOl lt u Just one such root s existence is guaranteed by the lemmas provided qt pz t satisfies either Ll lqtOllqltotl lt u and qlt w i 0 whenever both lt7t0l lt u and lw7t0l lt u or L2 moow lt H and l1 7 qtwoo lt 1 whenever ier lt H Now as z gtzO and w gttO and t gttO we nd i qlat w plz we gt 8pm tome ee 0 ii ciao2720matte plane tOplz mo gt apes taBzBMZO twat iii qtOz7zOq39t plane tOapz tom gt 8pm taazamzo tome and iv 1 7 ll 7 q39tq39tOl l 7 ll 7 Bpz tBtBpzo tOEtl gt 1 Choose any positive constant K gt l3pzo tOBzBpzo tOBtl By continuity tiny positive tolerances 51 52 B3 54 ul uz M3 M4 must exist such that i qlat w plz awn ee 0 if z7zOl lt 151 amp m lt H1 amp w7toi s H1 ii iciao2720elem lt K if i27in lt 152 amp m lt H2 iii iqtOz7zOq39t i lt K if i27in lt 153 amp it 7 tot s H3 and 0quot l1Q39tQ39tol l1 313Z tVatBIKE to3tl lt 1 if izizoi lt B4 amp it toi 3 H4 Given any positive u lt minj uj keeping lz7zol lt B min mini 5 uK ensures that u satis es both lemmas requirements thus guaranteeing that just one root t exists in lt7t0l lt u End of proof of Implicit Function Theorem This theorem s proof works for arbitrary analytic functions pz t of two complex variables not just polynomials but it is not the best theorem It establishes the differentiability of simple roots of analytic equations without revealing what happens when roots coalesce In fact roots remain continuous if not differentiable where they coalesce we shall prove this when we come to Rouch s theorem For now we are content with the Implicit Function Theorem s assurance that every algebraic expression has a continuous divided difference and a derivative both also algebraic expressions except at Branch Points where roots coalesce and across somewhat arbitrary Slits introduced to make the expression singlevalued Like rational expressions algebraic expressions can have Poles where they take in nite values but poles need not detract from continuity and differentiability if the expressions are construed as maps from the Riemann Sphere to itself For this sphere see the notes on Mquotbius Transformations Elementary Transcendental Functions Nonalgebraic analytic functions are called Transcendental The Elementary Transcendental functions arise out of algebraic operations upon expz and lnz The latter s multiplicity of values each differing from others by integer multiples of 2m is suppressed by a notation that assigns one Principal Value to lnz in some contexts and if done conscientiously uses another notation like Lnz for the multivalued version One text Complex Variables and Applications 6th ed 1996 by Brown amp Churchill McGrawHill does just the opposite Prof W Kahan Page 9 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Formulas Defining Principal Values of Inverse Elementary Functions Complex 2 X 1y has 2 x 71y for real Rez x and Imz y 12 7l ix 1yl Wx2 y2 in other words lzl WzE 2 0 argx 1y 2 arctanyX ix 1yl if y i 0 or X gt 0 so 77E lt argx1y lt T signyl 7signxTE2 otherwise where sign il always eXpX 1y eXcosy 1siny lnz lnlzl 1argz This principal value has 77E S Imlnz S T zW expwlnz except 201 for all z and 0W 0 if Rewgt0 Wz 212 This principal value has Rez 2 0 arctanhz lnlz 7 lnl7z 2 7arctanh7z arctanz arctanhlz1 7arctan7z arcsinhz lnz Vl 22 7arcsinh7z arcsinz arcsinhlz1 7arcsin7z arccoshz 7 2ln Wzl2 kn2 arccosz 2ln Wlz2 1Vl7z2 1 TE2 7 arcsinz Exercise 9 Locate the locus in the complex zplane of each formula s discontinuities if any These formulas discontinuities their slits are located in the most commonly expected places Also in accord with consensus are the values taken on the slits acquiescence to the convention sign0 1 is tantamount to attaching each slit to its side reached by going counterclockwise around its one nite end But this C ounter Clockwise Continuity is too simple a rule to work for a function whose slit is a finite line segment Consequently some of the usual definitions of arcsecz arccoslz arccscz arcsinlz arccotz arctanlz arcsechz arccosh lz arccschz arcsinhlz and arccothz arctanhlz may change one day as arccot did it used to be arccotz TE 2 7 arctanz until about 1967 but now its slit is a finite line segment joining logarithmic branchpoints at z i1 and poked at z 0 The usual definition of arcsechz violates counterclockwise continuity around 2 0 These and many other annoying anomalies like Vlz lVz and argE 7argz just when 2 lt 0 go away when a signed zero is introduced though it brings a new anomaly many people nd more annoying namely that 74 10 74 7 10 but 21 W 74 10 i W 71 7 10 721 This is treated along with other perplexing examples and numerically stable algorithms for the formulas above in my paper Branch Cuts for Complex Elementary Functions or Much Ado About Nothing s Sign Bit pp 165211 in The State ofthe Art in Numerical Analysis 1987 ed by A Iserles amp MJD Powell for the Clarendon Oxford Univ Press Divided differences of transcendental functions cannot be simplified to eliminate the division z7w without incurring an integral see Hermite s formulation above Names have been given to some instances like explz 7z sinhzz and lnllz l7z arctanhzz and formulas like tanlz w l tanwtanztanz7wz7w often attenuate roundoff Exercise 10 Verify from the definitions of exp and In above that they are complex analytic functions because their real and imaginary parts satisfy the CauchyRiemann equations Find a short algebraic not transcendental expression for the derivative of In and of each arc Prof W Kahan Page 10 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Harmonic Conjugates A complex function fz of a complex argument 2 is called Analytic when it is complex differentiable on an open domain in the zplane We have seen that such an fz decomposes into real and imaginary parts that must satisfy the CauchyRiemann equations on its domain fX1y gx y 1hx y 3gx y3X 3hX y3y 3gX Way 43W y3X In other words the CauchyRiemann equations are necessary for analyticity They are su icient too because whenever they are satis ed by given functions g and h these define a complex function f g 1h whose 2vector interpretation s derivative f is one of the special 2by2 matrices isomorphic with complex numbers then the complex derivative is f39 BgBx IBhBx BhBy 713gBy What if g is given but not h Can we determine whether g is the real part of an analytic function f and if so then recover f g h from g Yes and yes to a degree Wherever g and h satisfy the CauchyRiemann equations in some open domain they are both Harmonic Functions therein because each must satisfy Laplace s Equation BZgBXZ BZgByz 7 0 and aZhBXZ 32M8y2 7 0 These equations follow from the CauchyRiemann equations and the observation that the order of differentiation can be reversed BBgBxBy BBgByBx provided the derivatives are all continuous This proviso will be assumed here even though it could have been deduced instead Consequently only harmonic functions are eligible to be the real parts or the imaginary parts of complex analytic functions The imaginary part h of a complex analytic function f g h is called a Harmonic Conjugate not complex conjugate of the real part g Both of them are harmonic in some open domain wherein they satisfy the CauchyRiemann equations Then g is a harmonic conjugate of 7h not h Either g or h determines the other minus an arbitrary real constant either determines f minus an arbitrary real or imaginary constant as we shall see next Books exhibit several ways to recover an analytic fx1y gx y 1hx y from a given harmonic gx y Most textbooks do it this way Define Hx y l BgBx dy and then obtain hx y Hx y 7 l BHBx BgBy dx 7 C for an arbitrary constant C The claim is that f g 1h is analytic To justify the claim we need merely verify that g and h satisfy the CauchyRiemann equations BhBx BHBx 7 EHBx BgBy 7BgBy as it should and ahay 7 aHay 7 a i aHax agay dx ay BHBy 7 l BBHBxBy EizgBy2 dx if all derivatives are continuous 7 aHay 7 i 3aHayyax aZgay2 dx BgBx 7 l BBgBxBx EizgBy2 dx BgBx as it should because 232g2x2 EizgBy2 0 The integrals above have been written as Inde nite Integrals to hide an arbitrary constant that lurks within h Another way to deal with that constant is to choose a point x0 yo inside the domain where g is harmonic and arbitrarily set hxO yo 0 this is tantamount to using Definite Integrals rst H ldy running from x yo to x y and then h H 7idx Prof W Kahan Page 11 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am running the integration from x0 y to x y But then these paths of integration must stay within the domain wherein g was given harmonic thus restricting the recovery of h and f glh to whatever subregion of the domain is reachable by such paths from x0 yo For example if x0 yo is centered in a narrow rectangular domain whose edges make angles of iTE4 with the real and imaginary axes the reachable subregion is a small hexagon Exercise 11 Explain why Of course x0 yo may be moved around the domain to reach other parts of it but then h may change by some additive constant Must all these changes be consistent with one function h over the whole domain Not necessarily Exercise 12 Except at the origin in the x 1yplane gx y lnx2 y2 is harmonic Its harmonic conjugate can be recovered by using the recipe above do so and show that no choice of constants yields one harmonic conjugate h continuous throughout the whole domain of g Another way to recover h and f from a given harmonic g can be found in many textbooks they use Green 3 Theorem in the Plane iaR Pdx Qdy UR BQBx 7 BP By dx dy wherein Px y and Qx y are continuously differentiable functions and R is a plane region whose boundary BR is a piecewise smooth closed curve traversed during the rst integration in a direction that puts the interior of R on the left Green s theorem is the attened into two dimensions version of the threedim ensional Stakes Theorem 3R v dr R curlv n dR wherein v is a continuously differentiable 3vectorvalued function of position in a 3dimensional vector space R in this space is a smooth surface whose edge is a piecewise smooth curve BR traversed for the first integral in infinitesimal steps dr and n is the unit normal at a point on R where dR is the infinitesimal element of area during the second integration oriented according to the RightHandRule viewed from the traversal of BR Green s theorem is often used to prove Stokes To obtain Green s from Stokes choose for v a vector in the same plane as R with components P and Q therein thereby ensuring that curlv be parallel to the plane s n Given any harmonic function g a continuously differentiable solution of Laplace s equation on some open domain set P iBgBy and Q BgBx into Green s theorem to infer that the integral JBR iBgBydx BgBxdy HR B2gBx2 B2gBy2 dx dy 0 around the boundary BR of every piecewise smoothly bounded subregion R inside the domain of g Therefore we may select any nite point x0 yo inside that domain and de ne hx y i49gBy39dx BgBxdy integrated along any piecewise smooth path from x0 yo to x y that stays strictly inside that domain This hx y is de ned independent of the path PROVIDED every two such paths with the same endpoints x0 yo and x y enclose between them only points interior to that domain wherein g satis es Laplace s equation the points between the paths constitute the subregion R for Green s theorem which implies equality of the integrals along both paths Prof W Kahan Page 12 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am We shall explore soon that proviso about points between the paths rst what does between mean A point lies between two paths just when every ray from that point to co crosses the paths taken together as one closed curve an odd number of times counting the curve s self crossings as multiple raycrossings two for an X selfcrossing three for four for 5 etc How do we con rm that this h defined as a pathindependent integral is truly a harmonic conjugate of g We compute the partial derivatives of h For this purpose we extend the path of integration along a line parallel to one of the coordinate axes Parallel to the xaxis dy 0 and consequently Bhx yBx Bi7BgBydxBx QgBy similarly BhBy BgBx so g and h satisfy the CauchyRiemann equations as conjugates should Since our pathindependent de nition of h says hx y Im RagBx 7 13gBydx 1dy it de nes an analytic function fx 1y gx y lhx y as a pathindependent integral fz if39zdz RagBx 7 13gBydx 1dy of its derivative f39x 1y 23ng 7 lagBy to within an additive constant In fact every analytic function is the pathindependent integral of its derivative so long as paths are restricted to the interior of the function s domain This is not the case for every differentiable real function of a real variable some real derivatives oscillate too violently to be integrated Later we shall learn every analytic function s integral is path independent too so long as paths and all points between them stay inside the function s domain Paths pose problems when at least one point not in the domain lies between them At the cost of oversimplifying the subject the problems can be dispelled by restricting attention to Simply Connected domains in the plane these are domains without holes Other characterizations of such domains include Whenever two paths inside the domain have the same endpoints all points between the paths lie inside the domain too Every closed curve inside the domain can be shrunk continuously all the while remaining inside the domain to a point inside the domain The last characterization also characterizes simply connected domains of dimensions higher than 2 and these can contain bubbles but not holes for example a cantaloupe s edible part is simply connected but a donut is not On any simply connected open domain in the plane every function g harmonic on that domain has a harmonic conjugate h de ned uniquely but for an additive constant everywhere on that domain by the foregoing pathindependent integral The construction of a unique but for an additive constant harmonic conjugate h of g is the attened version of the unGrad operation upon an irrotational flow in Euclidean 3space If qv is a continuously differentiable 3vector function of position v in a simply connected 3dimensional domain whereon curlq 0 then qv grad v for some scalar Potential function v unGradq I qv dv independent of the path of integration Exercise 13 How should q be determined by g to get 0 h 7 Thus by applying either Green s Theorem or the unGrad operator to the derivatives of a harmonic function g we may recover its harmonic conjugate h uniquely to within an additive constant as a pathindependent integral of the derivatives of g provided all points between paths lie inside the domain of g The proviso is necessary for some harmonic functions as in Exercise 12 but not all Prof W Kahan Page 13 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Exercise 14 Prove that gx y xx2 y2 is a harmonic function whose conjugate is obtainable from a pathindependent integral despite that its domain is not simply connected Exercise 15 On a simply connected domain consisting of the whole x yplane except for a slit cut along an arbitrarily chosen smooth Simple not selfintersecting curve joining 0 to co the harmonic function gx y 1nx2 y2 has a harmonic conjugate hx y Show that y 7 xtanhx y2x ytanhx y2 stays constant perhaps 00 throughout the domain while h runs through a range of real values whose extremes cannot differ by less than 47E so no singlevalued expression of the form 2 arctan constant can match hx y Exercise 16 The Critical Points of gx y are the points x y where BgBx BgBy 0 they may be maxima minima or saddlepoints of g Show that conjugate harmonic functions have the same critical points Later we shall learn that none of these can be local maxima nor minima interior to the functions domain Exercise 17 The Level Lines of gx y are the curves in the x yplane on each of which g is constant Show that the level lines of two harmonic conjugates form a family of Orthogonal Trajectories one function s level lines intersect the other s orthogonally except at critical points Orthogonality alone does not imply harmonic conjugacy show that the level lines of 2x2 y2 and of y2x are orthogonal trajectories though neither function is harmonic much less conjugate However if two harmonic functions level lines form orthogonal trajectories the functions can be proved to be each a constant multiple of the other s conjugate Exercise 18 Suppose gx y gx 7y is harmonic on a domain that includes a segment of the real x axis in its interior deduce that g has a harmonic conjugate hx y 7hx 7y and therefore fx 1y gx y 1hx y is an analytic function that is real on that segment of the real axis This justifies the term Real Analytic Function for any complex analytic function fz that is real on a segment of the real zaxis strictly inside the domain of f though some extra work is needed to deduce that this f must satisfy fx 1y fxi 1y The simplest way uses Taylor Series and Analytic Continuation qv Every standard elementary function or if multivalued its Principal Value discussed in the Chapter IV of our text Notes on Complex Function Theory by Prof Donald Sarason 1994 is a real analytic function Exercise 19 By integrating derivatives of a harmonic function gx y we recover one of its harmonic conjugates hx y and then set fx 1y gx y lhx y to recover an analytic function fz The recovery of an analytic expression f from a harmonic expression g that is also a real analytic function of each of its arguments can be achieved more easily by setting fz 2gEOz2 1Eoiz 2 for any 20 x0 1yO inside the domain of g explain why If this procedure fails to recover a function f analytic throughout the domain of g explain why Analytic functions are simpler than harmonic conjugates because every analytic function f is the pathindependent integral fz fc JCZ f 39wdw of its derivative f 39 along every path inside the domain of analyticity regardless of whether all points between paths lie inside that domain regardless of whether the domain is simply connected We do insist that the domain of an analytic function be connected otherwise perverse things could happen like Exercise 20 Prove that if f 39 0 throughout its domain but this domain is not connected then analytic function f stays constant in each connected component of its domain though perhaps a different constant in a different connected component of that domain regardless of whether the component is simply connected Prof W Kahan Page 14 z 0 p ltgtltgt and the sign of n becomes arbitrary So far complex arithmetic imposes no impediment Now let uz be a smooth real function of position 2 x1y in the Euclidean plane No complex number u39z can serve as derivative because real duz u39zdz Instead we define Vuz Bux1yBx 1Bux1yBy to be u s complex Gradient so that duz Re VuzE Here Re and conjugation are the nuisances in icted by mixing complex arithmetic with Euclidean vectors Exercise 21 Confirm that Vuz is normal J to u s level line through 2 that Vuz points in the direction of infinitesimal motions dz that maximize duldzl and that this maximum is 1Vuz1 Use Green s theorem p 12 above to prove JBRWdz lHR Wizudxdy where the Laplacian Wizu BZusz BZuBy2 involves 2nd derivatives assumed continuous in an open region R Generally lm l Wdz depends on the path of integration Next let uz and vz be smooth real functions of position 2 x1y but not conjugate harmonic functions of x y so that wz uz 1vz maps the Euclidean plane into itself but not conformally Neither w39z nor Vwz provides a complex number to serve as the derivative of w which like the 2by2 matrix f39 on p 1 has not two but four real elements Still complex multiplication of V and w yields an interesting object Vw BBx 1BByu 7 iv V w 7 I39VXW combining the Divergence V w BuBx BvBy with the Scalar Curl Vgtltw BvBx 7 BuBy Do you see why W vanishes when wz is an analytic function of z 7 Otherwise Vw accounts for the pathdependence of JC Wzdz which need not vanish if C is a loop as follows Exercise 22 Confirm a complex analog 13R wzdz lHR Vwdxdy of Green s theorem by applying it twice Verify that wdz W dz 1wgtltdz W t 1W n1dzl where t is the unit tangent to BR so directed that R lies on the left and n 71t is the outward right pointing normal to BR Then the complex Green s theorem yields Stokes theorem 13R W t1dzl l R Vgtltw dxdy and Gauss Divergence theorem 13R wmldzl R V wdxdy in the plane thus condensing two renowned theorems into one cryptic equation 13R Wzdz lHR Vwdxdy Into the last equation substitute w gVh where gz and hz are smooth real functions of position 2 x1y to get laRgWdz lHR VgW glVlzh dxdy This will figure in the characterization of harmonic functions as solutions of the following variational problem Suppose region R is inside the domain of a harmonic function h so Wizh 0 in R and suppose at least part of the boundary BR does not cut levellines of h orthogonally along BR s remainder if any h s normal derivative BhBn ReVhn lmmt lmmdzyldzl vanishes Except on that remainder suppose g 0 on BR Thus gh runs over smooth functions whose boundary values on BR match h there Dirichlet conditions except perhaps on that remainder where BhBn 0 Neumann conditions Of all such smooth functions the one that minimizes HR 1Vgh12dxdy turns out to be h because HR1Vgh12dxdy HR 1Vg12 1Vhlz 2ReVgW dxdy HR 1Vg12 1Vh12 7 ZglVlzhdxdy m IaRgWdz 711141ng 1Vh12dxdy 2 HR thlzdxdy with equality just when g 7 o in R In his thesis G Riemann took for granted that a minimizing harmonic h must exist for any given piecewise smooth boundary values on BR Prof W Kahan Page 15 Notes for Math 185 Derivatives in the Complex zplane May 16 2008 413 am Summary of the Next Few Topics 1 Every analytic function f is the pathindependent integral fz fc If f39wdw of its derivative f 39 along every path inside the domain of analyticity regardless of whether all points between paths lie inside that domain regardless of whether the domain is simply connected Proof Green s Theorem or unGrad exploit the CauchyRiemann equations and since f and the integral of its derivative have the same derivative throughout the domain they must differ by a constant thereon 2 Every continuous function f whose integral Fz lcz fwdw is pathindependent on some open domain connected to the point c is the derivative fz F39z of its integral which is therefore analytic on that domain Proof FlzAzz 7 fz gt 0 as lAzl gt 0 3 The integral l fzdz of every analytic function f is pathindependent and therefore analytic on every simply connected open subset of the domain of f CauchyGoursat theorem 4 Cauchy s Integral Formula fnz n lC fwwiz 1 ndw27E1 for every integer n 2 0 if z is inside a simple closed curve C inside which f is analytic on and near which f is piecewise continuous so all derivatives fnz exist and are analytic too Goursat s proof 5 Every continuous function whose integral is pathindependent throughout some domain it is evidently connected but perhaps not simply is analytic thereon Morera s theorem 6 Wherever f is analytic so is fl since fl gt f39 This is an example ofa Removable Singularity If inside an open domain F is analytic everywhere except perhaps at one interior point around which F is bounded then F can be redef1ned at that point to render it analytic there too Riemann s removal of a singularity 7 The only bounded entire functions are constants Liouville s theorem 8 Every complex polynomial has as many zeros as its degree counting multiplicities Gauss 9 Every analytic function equals its average value on a concentric circle Gauss 10 The Taylor Series of an analytic fz 2120 2720n fnzOn converges absolutely and is termbyterm differentiable and integrable within its circle of convergence on which lies the singularity of f nearest zO The series diverges outside this circle At any point on this circle where the series converges it converges to the nontangential limit of f Abel s theorem 11 Every analytic function s magnitude takes its maximum value over its domain somewhere on its boundary Every nonconstant analytic functions maps interior points of its domain only to interior points of its range Maximum Modulus Theorem Open Mapping Theorem 12 Every analytic function s magnitude takes its minimum value over its domain somewhere on its boundary if not at its zeros inside the domain D Alembert s principle 13 If f and g are analytic throughout the same domain and if l f 7 gl lt lgl on its boundary then f and g have the same number of zeros inside that domain Rouch s Theorem Prof W Kahan Page 16 MATH 185 TOPOLOGY WORKSHEET Dave Penneys Summer 2009 1 Open and Closed Sets Recall the following de nitions from class De nition 11 The 8 ball or e neighborhood around 20 E C is Bg20 z E Cllz 7 20 lt 8 The deleted 6 ball or deleted e neighborhood around 20 E C is Bg20 20 z E C0 lt12 7 20 lt 8 De nition 12 Let S C C A point 20 E C is called 1 an interior point of S if there is an 8 gt 0 such that Bg20 C S7 2 an exterior point of S if there is an 8 gt 0 such that Bg20 C S07 and 3 a boundary point of S if 20 is neither an interior point nor an exterior point of S Exercise 13 Show that 20 is a boundary point of S if and only if for all 8 gt 07 3520 m S 7 0 7 3520 D So De nition 14 Suppose S C C We de ne 1 the boundary of S as 3S z E Clz is a boundary point of S7 2 the interior S as intS S 3S7 and 3 the closure of S as S S U 3S Exercise 15 Show that 3S 3S0 De nition 16 De ne the exterior of S as extS Sc 3S Exercise 17 Show that extS intSc De nition 18 S C C is called 1 open if S intS7 and 2 closed if S S Exercise 19 Show that S is open if and only if S0 is closed Hint Use Exercise 15 and DeMorgan7s Laws Exercise 110 Show S Hint Use Exercises 17 and 19 Exercise 111 True or False 1 30 2 intSC intSc Exercise 112 Show that 1 S intScc7 and 2 intS S0 De nition 113 20 is an accumulation point of S if each deleted e neighborhood of 20 intersects S7 ie for all 8 gt 07 Exercise 114 Find a set with 1 no accumulation points7 2 one accumulation point7 and 3 n accumulation points where n E N 2 Compactness Theorem 21 Hiene Borel The following are equivalent for a subset S ofR 07quot C39 I S is closed and bounded 2 Every open cover ofS has a nite subcoueiquot ie if Uili E I is some collection of open sets such that S c U U i39eI then there is an n E N such that SCUi1UUUi De nition 22 A subset S of R or C is called compact if one both of the conditions in 21 hold Exercise 23 Use the second condition in 21 to show that the continuous image of a compact set is compact7 ie7 if S C C is compact and f C a C is continuous7 then fS is compact Hint Note that f is continuous if and only if f 1U is open for every open U C C Theorem 24 Extreme Value Theorem Suppose U C C is open f U a R is continuous and S C U is compact Then there are 2021 E S such that for every 2 E S fzo S f2 S f217 ie f attains its maacimum and minimum on S De nition 25 Suppose S C C and 20 E C Then distz7 S inf 7 zollz E S Exercise 26 Suppose S C C Show that f C a R20 given by fz distzS is continuous Theorem 27 8 Collar Suppose U C C is open and K C U is compact Then there is an 8 gt 0 such that the open set Us z 6 Cl distzK lt 8 C U Obviously K C Us Proof Look at the function f C a R20 given by fz distzUc As f is continuous7 by the Extreme Value Theorem 247 f achieves its max and min on K As K U0 7 we have that the minimum must be strictly greater than 07 ie7 there is an r gt 0 such that distzUc 2 r for all z E K It 0 lt 8 lt 7 then B42 C U for all z 6 K7 so 2 6 Cl distzK lt 8 U B42 C U 26K 3 Paths and Contours De nition 31 A path in C is the image of a continuous map 7 a7 b a C where a lt b E R A path 7 is called linear if yt 1 7 t20 t21 for 2021 E C In this case7 y parematrizes the straight line from 20 to 21 closed if 7a yb simple closed if y is closed and if t17t2 6 ab with t1 31 t27 then yt1 31 yt2 ie WW is injective7 smooth if y is continuously differentiable on ab and y t 31 0 for all t 6 a7 b7 a contour if y is a piecewise smooth path Exercise 32 Let y be a path Deduce im y is compact Exercise 33 Glueing Lemma Suppoe 71 17 a C and 72 bc a C are two paths such that 71b 720 Show 73 a7 c a C is a path where 7 71t if t 6 ab 73 Wm if t e bu We usually call 73 the concatenation of 71 and 72 usually denoted yg 71 72 3 Exercise 34 Suppose W 17 a C is a path Show W 4977a a C is a path where Wt W7t We usually denote W 7W De nition 35 A path W is called piecewise linear if W is the concatenation of nitely many linear paths Exercise 36 Suppose W 1 b a U is a path where U C C is an open set Show that there is a piecewise linear path W ed a U such that Wa and Wb Hint Use the 6 Collar Theorem 27 De nition 37 A path W 17 a C is called horizontal ifW is linear and lmWa lmWb7 vertical ifW is linear and ReWa ReWb7 and rectangular if W is the concatenation of nitely many horizontal and vertical paths Exercise 38 Suppose W 1 b a U is a path where U C C is an open set Show that there is a rectangular path W ed a U such that Wa and Wb Hint Use the 6 Collar Theorem 27 4 Path Connected and Simply Connected Sets De nition 41 A subset S of C is called disconnected if there are nonempty open sets UV C C such that 1 U m V 0 2 S U7 7 S V7 and 3 S C U U V A subset of C which is not disconnected is called connected Exercise 42 Show that Q is connected De nition 43 A subset S of C is called path connected if for all 20 21 6 S7 there is a path W 17 a S such that Wa 20 and Wb 21 Exercise 44 Show that path connectedness implies connectedness Example 45 De ne P 000 a C by Ht t isin Let S imP U Then S is connected7 but not path connected There is no path W 17 a S with Wa 0 and Wb 1 Theorem 46 Suppose U C C is open Then U is connected if and only if U is path connected De nition 47 A subset D C C is called a domain if D is open and connected De nition 48 A subset S of C is called bounded ifthere is an N E N such that S C BN0 Theorem 49 Let y be a simple closed path in C Then im yc is disconnected Moreover there are connected open sets U V such that 1 U m V 0 2 im yc U U V and 3 U is bounded and V is not bounded We denote U ins y and V out y the inside and the outside of y De nition 410 A domain D is called simply connected if for every simple closed path 7 ab a D we have insD Dc 0 Exercise 411 Show that C 0 is a domain but not simply connected Exercise 412 Show that the continuous image of a connected set is continuous 5 Local Uniform Convergence De nition 51 Suppose fn is a sequence of functions f a C for some domain D and suppose f D a C 1 If S C D we say fn a f uniformly on S if for all e gt 0 there is an N E N such that lfnz 7 lt 8 whenever it 2 N and z E S 2 We say fn a f locally uniformly if fn a f uniformly on every compact subset of D Fact 52 If fn a f uniformly then fn a f locally uniformly Exercise 53 Show that if D B11 fz 17 z 1 and 70 then fn a f locally uniformly on D but not uniformly 1 n71 Zn H t N t th t 739 in Ge aliz ltzgt1z Theorem 54 Suppose fn a f locally uniformly on some domain D I If fn is continuous for all n E N then so is f 5 2 Suppose fn is continuous for all n E N If y cub 7 D is any contour and g y 7 C is continuous then gin fn292 dz 7 f2d2 dz 3 If fn ls holomorphlo for all n E N then so is f Proof 1 Let 20 6 D7 and let 61 gt 0 such that K B5120 C D Let 8 gt 0 Pick N E N such that lfNz 7 lt 83 for all z E K Pick 62 gt 0 such that lfNz 7 fN20l lt 83 for all l2 7 20 lt 62 Setting 6 min51527 we have that whenever l2 7 20 lt 6 8 lf2 f20l S lf2 fNZl lfNZ fNZol lfN20 f20l lt 33 8 2 As 9 is continuous on the compact set im y7 by the Extreme Value Theorem 247 there is an M 2 0 such that S M for all z E im y Let L be the length of y and let 8 gt 0 Pick N E N such that 7 lt 8ML for all n 2 N and all z E im y Then by the lntegral Bounding Theorem7 fn292 dz7 f2d2 dz 7 fn27f2d2 dz lfn27f2lld2l dz 8 LiM lt ML 8 3 Suppose y is a simple closed contour such that ins y C D By part 2 and the Cauchy Goursat Theorem7 we have fz dz lim dz lim 0 0 7 7 As f is continuous by part 17 by the stronger version of Morera7s Theorem7 we are nished FilezMobius Notes on Mobius Transformations September 29 2006 647 pm Linear Fractional Bilinear Rational Mobius Functions The function Mz uz Bb 7 mz is determined by the values of four constant parameters 5 u b and m chosen almost arbitrarily subject to an inequality constraint 9 Bm by i 0 that prevents Mz from degenerating to a constant function Actually just three parameters su ice to determine M since dividing all four of B u b and m by the same nonzero constant and dividing c by its square leaves M unchanged Therefore nothing important is lost by assuming that one parameter has the value 1 but we shall do this only when it is convenient The equation w Mz can always be solved for z Ww bw 7 Bu mw the two Bilinear7Rational functions M and W are inverse to each other in the sense that MWw w and WMz z Their parameters determine the same nonzero value c which appears in the Bilinear equation b 7 mzu mw c This is more symmetrical than equations w Mz and z Ww but equivalent to them only if m i 0 Differentiate it to nd 9 in the derivatives M39z 7 eb 7 mz2 and W39w 7 cp mw2 which vanish only at 00 if anywhere Evidently no bilinearrational function has a C ritz39calPoz39nt a nite place where its derivative vanishes otherwise it could have no inverse Recall that an analytic function cannot be a onetoone bijectz39on in the neighborhood of its critical point The foregoing expressions for their derivatives show that Real bilinearrational functions those whose parameters and argument are all real are all monotonic except across their poles and thus fall into two classes those that are increasing c gt 0 and those that are decreasing c lt 0 functions of their real argument Nothing quite like that segregates complex bilinearrational functions each can be treated as a point on a connected 3dim ensional quadric hypersurface embedded in a complex 4dimensional B u b mspace by restricting c Bm bu l say Restricting c i1 treats real bilinearrational functions as points on two 3dim ensional hyperboloids embedded in a real 4dim ensional B u b m space the two hyperboloids have no nite intersection though they approach each other arbitrarily closely out towards in nity where lie functions nearly constant nearly everywhere Exercise 1 Show that all bilinear rational functions M satisfy the same dilTerential equation 2M39M39quot 3Mquot2 all of whose other solutions are constants Hint Consider UNWquot Stereographic Projection and the Riemann Sphere The natural domain for bilinear rational functions like Mz uz B b 7 mz and its inverse Ww bw 7 Bu mw includes a single point at 00 Moo 7um W7um oo Woo bm and Mbm oo Adjoining this point at 00 to the complex plane C turns it topologically into a spherical surface S called the Riemann Sphere Let us imagine S to be a sphere of radius 1 centered at the origin in a 3dimensional Euclidean space of points w 9 in which 2 is the altitude above or below a copy of the complex plane in which w runs Then w2 22 1 when w 9 lies on S StereographicProjection identi es w 9 on S with a complex 2 in C by joining them with astraight line through 0 l at the North pole of S This maps 2 in C to Sz 22z2l z27lz2l on S and maps w 2 on S to Cw Q wl7Q in C whose unit circle projects to itself unchanged as S s equator Exercise 2 Con rm that the three points 0 l 82 and z 0 are collinear in w Q space and that the functions C and S are inverse to each other and that they associate the North pole 0 l of S with 00 in C the South pole 071 of S with 0 in C and S s equator w 0 on which wl l with the unit circle 2 w in C Stereographic projection takes two nite points 21 and 22 in C to antipodal points on S if and only if 212271 prove this Prof W Kahan Extracted from longer notes for Math 185 Page 16 FileMobius Notes on Mobius Transformations September 29 2006 647 pm Exercise 3 Show that Rewdw 79d when w 2 moves on S so moving 2 Cw 9 that dz dw179 wdQli f satis es 1sz ldwlz d9217 22 Show the Choralal distance between projections on S ofpoints 21 and 22 in C to be 2122721l121121122121 In many a text the Riemann Sphere S is a sphere of radius 12 resting on top of C at its origin this would halve the chordal distance and alter some of the algebraic details but not change the gist of our reasoning about S A crucially important property of stereographic projection is that it maps circles on S to and from circles and straight lines in C To establish this property we characterize circles on S as intersections with planes in w z space The equation of a plane 1 in w Q coordinates is Recw OLQ B 0 for an arbitrary choice of complex constant c and real constants 0L and B The distance from the origin 0 0 to l is MPH 11BlOLZ lc12 and must not exceed 1 if I is to intersect S where lwl2 22 1 The points w 9 in the circle where S and l intersect projectto z Cw Q in C so w 2 Sz 22lzlzl MAD121244 which when substituted into P s equation yields the equation 2Recz 1021271 5021244 0 that 2 must satisfy If 0LB 0 this is the equation of a circle in C whose radius squared turns out to be cc2 1c121 7 HPHZOLB2 2 0 but if OLB 0 it is the equation ofa straight line in C projected through 1 to its intersection with S a circle through 0 1 in w Q space Exercise 4 Con rm the previous four sentences assertions Where are the circles centers Stereographic projection is Conformal it preserves curves angles of intersection To see why this is so consider two curves in the complex plane C that intersect at some angle 0 This is the angle between our curves tangents two straight lines that project to two circles on S tangent to the projections onto S of our two curves The two circles are cut from S by two planes each a plane through one of the tangent lines in C and through the North pole of S Here the circles intersect at the same angle as do their tangent lines which are the two planes intersections with another plane parallel to C through the North pole that angle is 0 again The circles other intersection also at angle 0 is at the projection onto S of our curves intersection point so the projections onto S of our curves intersect at angle 0 too Proof due to C Caratheodory Every bilinear rational function Mz Hz Bb 7 mz is construed simultaneously as a map of the Completed at 00 complex plane C to itself and as a map of the compact Riemann Sphere S to itself thus w 9 maps to SMCw Q Both ways every such map is called a Mobius Transformation As we shall see the set of all Mobius transformations forms a non Abelian noncommutative Group under composition and every member of this group maps circles to circles on the Riemann Sphere S and circles to circles in the completed complex plane C subject to the understanding that if a circle passes through 00 it is actually a straight line in the plane Conversely any map of one simply connected open region on S to another that maps all circles inside the rst region to circles inside the second must be a Mobius transformation or its complex conjugate C Carathe39odory proved this in 1937 without presuming the map s continuity His proof will not be reproduced here alas Exercise 5 Extend analytic functions domain C to S as follows Show that every nite ordered pole of an analytic function f is a removable singularity and a zero of 1 f and that if Sfz approaches a pathindependent limitpoint on S as z gt oo then f1z has a pole of nite order or a removable singularity at z 0 Hint Recall the CasoratiWeierstrass theorem Prof W Kahan Extracted from longer notes for Math 185 Page 26 FilezMobius Notes on Mobius Transformations September 29 2006 647 pm The Completion of C by adjoining one point at ltgtltgt 10 is a mixed blessing On one hand it identi es C with a compact set S on which all continuous functions are uniformly continuous and all rational functions not merely continuous but differentiable even at their poles which seems advantageous On the other hand it can render formerly innocuous algebraic operations risque For instance a cancellation law zz l formerly violated only by z 0 is now violated also by z ltgtltgt as are other cancellation laws z 7 z 0 and zO 0 Do not confuse these operations invalid upon numbers when ltgtltgt is treated as a number with operations that look the same in print but are actually operations upon functions zz l and z 7 z 0 an zO ltgtltgt 0 are valid operations upon analytic functions after removable singularities at z 0 and at z ltgtltgt have been removed This removal is a limiting process not an arithmetic process that computers can carry out with mere numbers For this reason computers but not human mathematicians require some other symbol besides ltgtltgt to be adjoined to completions of the real and complex number systems such a symbol is NaN standing for Not a Number Without it any convention that assigned numerical values like 1 to 00 and oooo or 0 to ltgtltgt0 and co 7 co would merely propagate the cancellation laws failures to other laws like the associative and distributive laws Examples like l 00 3 00 300 31 3 and ltgtltgt 372ltgtltgt 3oltgt 7 290 co 7 co 0 and 0 ltgtltgt0 oooo l would undermine our faith in computation as the rst example did in APL and the others do in MthCad With NaN and with conventions that assign it instead of something more confusing to otherwise invalid arithmetic operations and that propagate it through all algebraic operations and most but not all other operations the order in which cancellation associative distributive and other algebraic laws are applied during the evaluation of an unconditional no comparisons arithmetic expression can affect its result in only this wa If different orders can produce different results just two results are possible and one of them must be NaN For example one application apiece of the cancellation and distributive laws turn 1 7 lz into z7lz so they are the same analytic function after the latter expression s singularity at z ltgtltgt has been rem oved39 but this rem oval cannot be accomplished by mere evaluation which yields NaN instead of l for the latter expression at z ltgtltgt NaN and co offer no panacea Aside from Riemann s removal of isolated singularities by a limiting process no methods exist in general to deduce uniquely correct extensions of familiar functions of real and complex arguments to cope with co and NaN This is partly because the completion of the complex plane 0 by a single point at ltgtltgt is not the only possible completion39 a line or circle at in nity might serve better in some circumstances Similarly the real line can be completed as well by two in nities ice as by one unsigned ltgtltgt Different completions generate NaN differently though they must always conform to the requirement displayed boldfaced above If computations had no way to get rid of NaN it would be useless computation would be better stopped before so persistent a NaN were created This means that some operations upon NaN must produce something else without exacerbating confusion For example NaN0 1 because 20 l by convention for all real and complex values of 2 including 2 0 and z ltgtltgt More generally suppose a function fz w is so de ned for all nite and in nite arguments that fz 0 is a constant independent of Z39 a plausible principle requires fNaN 0 to take the same constant value But this plausible principle can be ineffective For example predicate NaN 2 is True and z 2 False for all nite and in nite z 39 what should NaN NaN be It is True by de nition and so NaN NaN is False These astonish implementors of programming languages The exigencies of computer programming are driven sometimes less by logic than by accidents of history one of these forced NaN NaN to be declared True Another example Maxx ltgtltgt ltgtltgt for 7ltgtltgt S x S ltgtltgt implying that MaxNaN ltgtltgt eltgt too and suggesting that MaxNaN x is better de ned to be x than to be NaN even though NaN S x is False for all x as is NaN gt x Thus ltgtltgt that begets NaN may postpone or worsen complications not eliminate them Computer systems are expected to raise a Flag to signal the creation of co from nite arguments and to raise another ag when NaN is created instead of merely propagated Such a signal is obligatory because NaN and co can destroy the validity of programs designed without them in mind Programmers determined to prevent ltgtltgt or NaN from being created would have to scrutinize the inputs to every operation or subprogram that had not been proved incapable of creating these unwanted entities and then gure out what to do with possibly dangerous inputs Such scrutiny could be onerous tim econsum ing for both programmers and computers even if nothing dangerous ever materialized In most cases the speedier way on average is to check a ag afterwards and on the rare occasion if any when a NaN or ltgtltgt created unwittingly has corrupted computed results supplant these by something else Sometimes Topology annoys Algebra the open plane C is a Field the closed sphere S is not Sometimes Computing annoys Mathematics signal ags and NaN are mathematically ugly computational details Much as we yearn to avoid annoyances some of them are provably unavoidable and they put bread on many of our tables Prof W Kahan Extracted from longer notes for Math 185 Page 36 FileMob1us Notes on Mobius Transformations September 29 2006 647 pm The Group of Mobius Transformations Under composition Mquotbius transformations form a noncommutative group identi able with a quotient of subgroups of the multiplicative group of invertible 2by2 matrices To see how this works let MJz uJz Bjbj mJz and cj bJuj Bjmj 0 for j l 2 and 3 then see that for any such choices of M1 and M2 their composition M2M1z M3z is another Mquotbius transformation whose parameters can be computed by matrix multiplication thus b3 7m3 b2 imz b1 iml and c3 det b3 7m3 9291 i 0 53 H3 132 Hz 131 1 B3 3 Moreover the Mquotbius transformation M whose parameters are B u b m c bu Bm has an inverse W that may be assigned parameters respectively Bc bc uc mc lc just as 71 the inverse matrix r c But we cannot identify each Mquotbius transformation M H with one uniquely determined matrix since multiplying M s parameters B u b m c by a nonzero constant and c by its square changes the matrix without changing M Instead we can identify each Mquotbius transformation with a pair of matrices but rst we must restrict c If M is real restrict c to the set 1 71 if M is complex x 9 1 Doing so identi es each Mz uz Bb 7 mz with the pair ofmatrices iM i E m 11 whose determinants detiM bu Bm c The set of all such matrices includes their inverses and thus constitutes a noncommutative multiplicative group consisting of the quotient the group of all 2by2 matrices with determinants 1 or i1 its subgroup 1 7I This multiplicative quotient group is Isomorphic to the group under composition of Mquotbius transformations These groups factor naturally into the three subgroups tabulated here Table 1 Subgroups of Mquotbius Transformations Mz Subgroup Name Mz B u b m 9 Matrix I Dilation or Scalingl iuzz 0 0 i1 H 0 i1 idH 0 0 u Translation 2 B B l l 0 l 391 0 B 11 Inversioni 42 gt 1 gt 0 gt 0 gt 1 gt 1 gt 0 717 1 0 z 0 1 1 0 1 1 0 v 0 1 1 The i sign is only for Real Mobius transformations supplant it by for Complex The Inversion subgroup has two elements one of them the Identity Mz 2 that changes nothing Theorem Every Mquotbius transformation is a composition of at most four transformations each selected from a subgroup tabulated above and some Mquotbius transformations require all four Proof Mz uz Bb 7 mz 71m2cz 7 mbc 7 Wm is atranslation ofan inversion of a translation of a dilation unless m 0 in which case M is a translation of a dilation Either way no more than four selections from the subgroups su ice to reproduce M Three selections may be too few if m i 0 because then M requires an inversion to create the pole at z bm and then two more selections are too few to determine three parameters In other words three selections suf ce to generate only a nite number of twodimensional surfaces inside the three dimensional space of Mquotbius transformations End of proof Exercise 6 State the foregoing theorem in matrix terms and prove it those terms Prof W Kahan Extracted from longer notes for Math 185 Page 46 FilezMobius Notes on Mobius Transformations September 29 2006 647 pm Exercise 7 Verify that the Mquotbius transformations 2 lz 172 and all their compositions constitute a subgroup with six members the other three being ll7z l7lz and ll7lz Often an easy way to prove an assertion for all Mquotbius transformations is to do so separately for each tabulated subgroup and then infer it for their compositions Often the assertion is obvious for all but the Inversion subgroup Such is the case for the next three assertions oiTered as Exercise 8 Prove that Mquotbius transformations map circles on the Riemann sphere S to circles on S Any two given circles on S can be mapped one to another by Mquotbius transformations Mquotbius transformations map Re ections in a circle to re ections in its image circle Re ections in a circle are de ned thus Given the center c and radius r gt 0 of a circle in the complex plane C two points p and q are said to be re ections of each other in that circle just when Xq 7 c r2 or equivalently pwq7c gt 0 and lp7cl jqwl r2 Thus c p and q are collinear and the latter two are one inside and the other outside the circle but both on the same side of c For example when c Jr the re ection of p is q 50 7 lpr now x p while letting r gt 00 to nd that q gt 13 while the circle approaches the real axis In general two points are re ections of each other in a straight line when it is the right bisector of the line segment joining the points as one might expect Exercise 9 Show that all circles through two points each the re ection of the other in another circle 0 cut it orthogonally at right angles Hint Mquotbius transform 0 to the real axis Fixed Points The only Mquotbius transformation with more than two xedpoints is the identity which leaves all of S xed Every other Mz uz Bb 7 mz has one or two Fixed Points f Mf they are the zeros of the quadratic polynomial mz2 7 b7uz 5 subject to the understanding that 00 is a xedpoint just when m 0 When b7u2 7 4mB 0 two distinct points are xed call them f1 and f2 If both are nite M is de ned by a more symmetrical equation Mz 7 f1Mz 7 f2 kz 7 f1z 7 f2 all Mquotbius transformations M with the same two nite distinct xedpoints constitute a subgroup isomorphic to the multiplicative group of nite nonzero constants k the identity transformation has k l If one xedpoint f is nite and the other is 00 then Mz 7 f kz 7 f for the same group of constants k M has only one xedpoint f Mf just when b7u2 7 4mB 0 Ifthis one xedpoint f is nite then the more symmetrical equation is lMz 7 f K lz 7 f and the subgroup of such Mquotbius transformations is isomorphic to the additive group of nite constants K with K 0 for the identity If the one f 00 then Mz z K for the same group of constants K Exercise 10 Derive the more symmetrical equations from appropriate hypotheses about the quadratic namely that the xedpoints satisfy f1 f2 b7um and f1 f2 Bm when they are nite or appropriate other equations when one or both xedpoints are in nite Then use the more symmetrical equations to prove that a Mquotbius transformation M not the identity maps no circle onto itself on S just when M has two distinct xedpoints and lkl 1 and Imk 0 in this case M is called Loxodromic Otherwise a onerealparameter family of in nitely many circles each mapped onto itself by M covers S identify the three diiTerent kinds of family Prof W Kahan Extracted from longer notes for Math 185 Page 56 FilezMobius Notes on Mobius Transformations September 29 2006 647 pm Cross Ratios Z 7 p q 7 I Z er qep 39 If one member of the quadruple is co merely replace this crossratio by an appropriate limit which is tantamount to discarding the factors containing 00 All crossratios are nite and they are preserved by all Mquotbius transformations Mz uz B b 7 mz which means that Zip qir MZ Mp MQJ Mr Zir qip MZ Mr MD MQD Any quadruple z p q r of four distinct points on S determine a C ross Ratio holds for every such M Exercise 16 Con rm this equation for each subgroup in Table l and then for the whole group Exercise 17 Since there are 24 permutations of the points in a quadruple z p q r they could determine 24 crossratios but these take at most six values show that if X is one of those six values the others are lX 17X l17X lilX and llilX Hint Exercise 7 A Mquotbius transformation w Mz uz Bb 7 mz is determined completely by what it does to any three distinct values p q r of z It must map them to some three distinct values P Mp Q Mq R Mr respectively of w Given the triples p q r and P Q R we may construct M by solving a bilinear crossratio equation like 2 7 mm 7 rxw e RQ e P w e PQ e sz e no 7 p for either w Mz or its inverse z Ww thereby determining the constants B u b and m except for a common factor One member of the triple p q r can be 00 if the crossratio equation is replaced by an appropriate limit which is tantamount to discarding the factors containing 00 similarly for P Q R With that understanding we can rewrite the crossratio equation in a more symmetrical form wen Q R 7R 7P M q Zir qip W Q that exhibits explicitly the crossratios in question and also exhibits explicitly the construction of M as a composition of two Mquotbius transformations the rst from z to that takes p q r to 0 1 00 respectively and the second from to w that takes 0 1 00 respectively to P Q R Given triples p q r and P Q R determine M but not its parameters uniquely To see why this is so suppose there were two Mquotbius transformations taking the rst triple to the second composing one with the other s inverse would create a transformation which with three distinct xedpoints would have to be the identity The triple of distinct points p q r determines uniquely a circle Q or a straight line in C through all three points likewise P Q R determines another circle MQ as well as the Mquotbius transformation M that maps one circular boundary to the other What does M do to the circles interiors It maps the interior of one either to the other s interior or else to its exterior how do we tell which The question seems moot when either triple of points lies in a straight line which occurs when either qrqp or QRQP is real Otherwise Exercise 18 Show why M maps the interior of circle Q to the interior of circle MQ if and only if the imaginary parts of qrqp and of QRQP have the same nonzero signs Exercise 19 Why is it worth knowing that provided p q r and P Q R are entirely nite pqqrrpPQQRRP b mpxb mqb mr2lt3 Prof W Kahan Extracted from longer notes for Math 185 Page 66 Mathematics 185 7 Fall 2009 7 Michael Christ Lecture 11 Monday 9212009 We begin the long road towards a rigorous proof of rigorous versions of Cauchy s theo rem 0 First midterm exam is on Wednesday Proof that U71 May if lf39ytl M for all t 6 ab We used the fact that f dz f Cm dz whenever F G are real valued func tions satisfying Cm for all z E a 1 together with a cute little trick Writing Mikey7f for some 9 E R we were able to reduce everything to facts about integrals of real valued functions See your notes for details 0 One of the highlights of any rst course in complex analysis Proof of Cauchy s theorem for rectangles Letting I fR f we found that for every n there is a rectangle R inside R whose diameter is 2 times the diameter of R such that I g 4 fRn 1 Unfortunately the uppew bound lemma gives us an upper bound for the right hand side whicxh is proportional to 4 2 so as n a 00 we seem to be losing 0 Let 2 be the center of R The sequence is Cauchy hence converges to some limit w Analysis can be de ned as that branch of mathematics in which limiting operations are used to solve problems Now use the hypothesis that f w exists Write fz fw z 7 wf w hwz the remainder hwz is very small in a precise sense as 2 a w by the de nition of differentiability The integral over R of the linear approximation fw 27 wf w is exactly zero This integrand obviously has a primitive 4 fRT hwz dz tends to zero as was shown in class using the upper bound lemma the diameter of R and what we know about the size of hw Therefore I is smaller than any positive number hence it must equal 0 1 0 Next we proved Cauchy s theorem for a disk This version applies to any closed contour piecewise C1 closed path in the disk not to one particular curve as in the preceding result The idea is to use the above result to construct a primitive we already know from 21 that fundamental theorem of calculus if f has a primitive in a domain then f7 1 0 for any closed contour at all in that domain We don t need the contour to be simple The primitive is constructed by integration along paths as in our proof that whenever f7 1 0 for every closed contour there exists a primitive The new observation is that when the domain is a disk any convex open set works equally well we don t actually need the hypothesis for every closed contour it s enough to have it for rectangles 1 Cauchy s theorem for a disk already leads to a host of results which shed immensely light on the nature of analytic functions These depend on a representation formula To derive that formula from Cauchy s theorem for disks and rectangles we ll need a slightly more general statement in which our integrand is allowed to possibly fail to be differentiable at one single point This will be discussed on Friday since we have an exam on Wednesday