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# Honors Calculus III MATH 2315

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This 9 page Class Notes was uploaded by Ila Haag on Monday September 21, 2015. The Class Notes belongs to MATH 2315 at University of Virginia taught by Philip Graber in Fall. Since its upload, it has received 43 views. For similar materials see /class/209571/math-2315-university-of-virginia in Mathematics (M) at University of Virginia.

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Date Created: 09/21/15

CLASS NOTES WEEK 1 1 COMPLEX NUMBER SYSTEM In this class we will use the following common number systems 0 N 12345 natural numbers 0 Z 0 i1i2i3 i4 integers 0 Q a E Zb E Zb 7E 0 rational numbers 0 R real numbers 0 C complex numbers Note N C Z C Q C R C C where C means is contained in77 or is a subset of7 These number systems can be characterized by operations under which the number system is said to be closed7 closed under closed under 7 closed under 7 gtlt R is closed under limits C is algebraically closed as we will explain below Before explaining where C comes from let7s consider an important property of Q namely that it is algebraically closed over linear equations of the form axb0 Algebraically closed means that if ab E Q then there exists a solution x E Q to the above equation Now try to solve the quadratic equation 2 7 2 0 This has no rational solution Q is not algebraically closed over quadratic polynomials However we can insert a number xi into our number system to get a new number system consisting of numbers of the form a lnE where a b E Q This number system has 7 gtlt ab i cdx aic bid a b c d ac 2bd ad bc ab i ab cid i aci2bd bciad Cd i 0272d2 7amp7de 0272d2 39 ls xi a number or just a placeholder In some sense there is no di erencel Now we know that even all the real numbers aren7t algebraically closed over all polynomials We still can7t solve for instance 2 1 0 So taking the above strategy lets make up a number called i V71 Then we get a new number system C de ned as CabiabER For a complex number 2 abi a is called the real part and b is call the imaginary part note these are both real numbersl We also write a Rez and b lmz 1 2 CLASS NOTES WEEK 1 We can perform operations on complex numbers like so ab2 i 0d2 aic bjd27 a b2e d2 ac 7 bd be ad2 ab2 ab2e7d2 ae7bd be7ad ed2ed2e7d202d2 02d2239 11 Examples Number 127 p 5 in the handout asks us to calculate 2 271723 7 22 72 7 22 7 2 13 7 22 71 7 322 7 22 72 2239 7 62 7 6 78 7 42 Number 157 p 5 asks if k is an integer7 show that file 17 4k1 4k2 717 4k3 Note that 2390 121 222 7123 7239 and 2394 42 722 771 1 So 24k 24 1k 1 and the rest follow immediately lntroduction to proofs Number 297 p 5 Prove that there is no rational number x that satis es 22 2 Well7 suppose z pq7 where p21 are integers We can write p and q in lowest terms7 so they don7t have a common factor But 102212 27 so p2 2212 Then p must be even7 so p 27quot for some integer r Thus7 422 22127 so 12 222 That means q is even7 too7 which means p and q must share a common factor This is a contradiction 2 7239 2 GRAPHING COMPLEX NUMBERS We can use a Cartesian plane to plot complex numbers lf 2 z 2217 then the point 272 on the Cartesian plane represents 2 We call it the compleco plane We can de ne the absolute value or modulus of a complex number by measuring its length in the complex plane That is7 if z a b2 then the modulus is given by lzl Va2 b2 Eg 39 1 1374219165 In general7 Z 0 and 0 if and only if z 0 The distance between two complex numbers 21 and 22 is 21 7 22 lol 0 21 Examples of Graphs Basic example a circle is given by z l2 7 20 7 More examples Example 2 on p 9 asks us to nd the set of points z satisfying a lz2l 7 12711 b l2 71 Rez1 To graph a7 note what the equation is saying geometrically 2 is equidistant from 72 and 1 This gives a vertical line through the point 712 One could also get this by doing some algebra lz2l2 271l2 means that ifzab2 then a22b2a7 12b2 or a24a4a272a17 or 6a737 or a 712 To graph b7 let 2 a 1 b2 this gives a712b2 a 17 or a 712 b2 a 12 or 4a bz7 or a b24 This gives us a sideways parabola CLASS NOTES WEEK 1 3 Number 7f asks as to graph lzi1l lz1l 7 Thus the distance from 1 plus the distance from 71 is a constant this gives an ellipse with foci at i1 Letting z x y we have lzillz 77 lz1l2 49714l21ll21l2 implying x712y2 49714l21lx12y2 so 49714l21l4x 0 or 49 4x 14lz 1 Then 492 392 16x2 196 12 yz 196x2 392 196 19612 Thus 180z2 19612 2205 which is now more clearly the equation of an ellipse 22 Complex Conjugate The eomplem conjugate of a complex number 2 a be is 2 a 7 in For example 715 7175 7239 38 Properties 21i22 72 2122 2122 The proof of the second property takes a little bit of computation Let 21 a1 he 22 a2 be Then 2122 0102 7 5152 1152 azbi 1102 7 b1b2 7 1le agbl On the other hand 2122 11 7 12 7 0102 7 b1b2 7 0152 a2b1z39 This completes the proof Additionally we have lt220gt 22 22 Rezi 2 272 1 m2 2 372 lzll5l 222l2 These facts can be used to solve problems such as rationalizing the denominator 21 21 2172 22 225 l22l239 In particular 1 2 Z 7 My so that the reciprocal of a complex number 2 is parallel to the complex conjugate of 2 Some more useful properties From Problem 9 p 13 if 2 0 is a real number then lrzl This is easy to see directly if z a M then lrzl 771 Tb2 7 2a2 b2 rxaz b2 From Problem 10 p 13 lRezl g and llmzl g We7ll prove the rst the second is similar lf 2 a in then lRele lalz a2 S a b2 M2 take square roots to get the result From Problem 17 p 13 if a1a2 an are real coef cients then the polynomial z alzn l agzn z A can has roots that come in conjugate pairs real roots are their own conjugates To see this just take the conjugate of the polynomial using the properties mentioned above 4 CLASS NOTES WEEK 1 3 VECTORS AND POLAR FORMS A vector in the plane is given by two coordinates We can think of complex numbers as vectors in the plane Thus complex number addition can be performed by the parallelogram law77 note this is not the normal meaning of parallelogram law Take two vectors form a parallelogram the sum of the two vectors is the resulting diagonal across the parallelogram starting from the tail vertex common to the two vectors Since complex numbers can be treated as vectors in the plane we have the triangle inequality for complex numbers lzi 22l S lzil There is another side of this inequality namely llzll lzzll S 121 i 221 Exercise deduce the second inequality from the rst Example 1 on p 15 asks prove that the three distinct points 2122 and 23 lie on the same straight line if and only if 23 7 22 c22 7 21 for some real number c For the if77 direction let7s assume 23 7 22 c22 7 21 for some real number c Let 20 22 7 21 so that 22 20 21 Then 23 C20 22 c 120 21 so 22 and 23 are both on the line r20 21 through 21 For the only if77 direction if 21 2223 are collinear then there exists 20 such that 23 r320 21 and 22 r220 21 Thus 23 7 22 r3 7 r220 7 2 22 7 21 Let c 732 and were done A vector can also be given by its magnitude and direction We already have a notion of magnitude for a complex number its modulus If we identify a useful notion of direction we can characterize complex numbers in this way Cartesian coordinates zy can be converted to polar coordinates r 9 by solving the following equations z rcos97 y rsin9 For a complex number 2 2 iy we have T and we de ne arg2 9 Note however that 9 is only unique up to integer multiples of 27139 For example the values of argi are 7T 7T 7T 77i27T7i47T 2 2 2 so we write 7T argi 5 21 k E Z So arg2 has in nitely many values We wish to pick a de nite value The customary choice is in 7n7T and we call this Arg2 this is called the principal branch of the argument of 2 Any branch of arg2 is discontinuous along its branch cut77 having a jump of 27139 there If you wish you can use the notation arg 2 to mean the branch taking values in 77 27f eg arg7r Arg Note arg0 never makes sense since literally any angle would work We can now write complex numbers in polar form 2 z iy rcos9 isin9 rcis9 We will quickly see that the notation cis77 is unnecessary Examples 7b on p 22 asks us to convert 2 73 3239 into polar form Well r2 32 32 18 so r 3x2 So 2 7 i which points in the direction 9 37r4 So 73 3239 3x2cis37T4 7d on p 22 asks the same about 2 72 7 2239 for which r2 12 4 16 so r 4 Thus 231 77 7 5239 which corresponds to 9 77r6 CLASS NOTES WEEK 1 5 Adding complex numbers in polar form is inconvenient It can be done7 however Problem 217 p 24 if 7quotcis rlcisdl 772cis1927 we wish to nd 7 and 9 in terms of 771772191192 We have r2 71 cos 91 r2 cos 922 71 sin 91 r2 sin 922 7 cos2 91 2771772 cos 91 cos 92 7 cos2 92 7 sin2 91 2771772 sin 91 sin 92 7 sin2 92 7 7 2r1r2cos 91 cos 92 sin 91 sin 92 7 7 7 2771772 cos 1 7 92 We also have 71 sin 91 r2 sin 92 tan 71 cos 91 7 r2 cos 92 That7s all I will attempt here Now for multiplication we have it much better if 21 r1cos 91 739sin 91 and 22 r2cos 92 739sin 92 then 2122 r1r2cos 91 cos 92 7 sin 91 sin 92 7sin 91 cos 92 cos 91 sin 92 r1r2cos 1 92 739sint91 92 Thus7 mutiplying two complex numbers is the same as multiplying the moduli and adding the angles 2122ll21ll22l arg 2122 argzL arg22 Case study the powers of 739 are 17717717 These are just rotations of 7T2 around the complex plane Division is the inverse of multiplication and thus works in an analogous way 21 7 T1 coswl 7 92 S1n91 7 92 2 22 2 arg 71 argzL 7 argzz 22 21 22 31 Examples Problem 7h asks us to write 7W1 239 V3 7 in polar coordinates The modulus of the top is v 14 The modulus of the bottom is 2 So the modulus of the quotient is 142 The principal argument of the top is 737T47 and the principal argument of the bottom is 7r6 So the argument of the quotient is 737r477r6 797r12 7 27T12 7117T12 14 So the answer is Tcos7117r12 7sin7117r12 Problem 11 on p 23 says7 using the complex product 1 75 7 73947 derive the formula 7r4 4tan 115 i tan 11239 6 CLASS NOTES WEEK 1 Using Cartesian coordinate multiplication we just get 1 25 724 1 225 7 102212 1 224 i102 1 2576 i 480239 i 100 1 239476 i 480239 476 i 480239 4762 480 956 i 42 Thus arg1 2 4 arg5 7 2 arg956 7 42 using the product rule for arguments In particular then n4 4tan 1715 tan l71239 Since tan is an odd function the conclusion follows 4 THE COMPLEX EXPONENTIAL In order to rigorously de ne the complex exponential function 61 we are going to have to borrow from more advanced analysis one way or another For instance one could simply accept from the the ory of complex differentiation that there exists a function fz such that f z Alternatively lets just de ne I ez em y emcos y 2 sin We can easily verify that this is its own derivative with respect to each of the variables x and 22 that is 16 Z39Eiy y d7cosy 2s1ny isiny 2cosy 2cosy 2s1ny 1 This function has all the familiar properties of exponentials 621622 61112 611 76 612 21722 We also have consistent Taylor series expansions 22 3 24 5 1 E 22 x4 cosx17 gi 23 x5 s1n2zig 7 Just take the power series for em and apply it at x 2y 2 3 4 2y nini i e 712y 2 23X4X 2 4 3 5 n 1 yin 39 1 7i 1 2W4 HG 3W5 cosy 2siny This relation is referred to as Euler s Equation Note these functions are in fact analytic on the entire complex plane so these power series equations are literally true Justifying this assertion however requires some more advanced complex function theory CLASS NOTES WEEK 1 7 Now we have a new shorter polar representation 2 rem where r and 9 arg 2 Notice that 62 is 2W2 periodic That7s because 62quot 1 for every k E Z Famous example 5 1 0 This equation relates ve of the most famous constants in mathematics Useful identities 0039 We 2 29 72399 sin9 lmew 6 76 22 T162191 T262192 T1T26i9192 w 7151 2 51101762 736192 r2 72399 rem re Another interesting identity is known as De Moivre s formula cos 9 239sin 9 cos 729 2 sin 729 This is easily proved using the facts about 62 we have already seen yew 620620 62001 times 6262020 Ema Now apply Euler7s Equation 41 Examples Problem 3c7 p 31 1 06 ew4f 66267 seam239 Problem 4b p 31 2 2239 2925M fmm i 2 26257r6 Problem 127 p 31 Use DeMoivre7s formula and the binomial formula to derive the following a sin 39 3 cos2 9 sin9 7 sin3 9 b sin 49 4 cos3 9 sin9 7 4 cos 9sin3 9 For the rst7 note that cos39 2 sin39 cos 9 2 sin 93 cos3 9 32 cos2 9sin 9 7 3cos 9 sin2 9 7 239sin3 9 Taking imaginary parts7 we get a For the second7 note that cos49 239sin49 cos 9 239sin 94 cos4 9 42cos3 9 sin9 7 6 cos2 9 sin2 9 7 4239 cos 9 sin3 9 sin4 9 Once again7 taking imaginary parts7 we get Problem 207 p 32 For 2 7E 17 we have the classic geometric series formula z 1 7 1 1z22z 271 8 CLASS NOTES WEEK 1 Let 2 6w cost 2sint9 Then we get by DeMoivre7s formula 1 cos cos 219 cos n0 2sin sin 261 sinnt9 62n19 7 1 6w 7 1 62n192 7 672n192 T 5202 7 57202 i 22 sinn 1 2 22sin 2 i sinn 1 2 T sin 2 Taking imaginary parts we get the formula 622202 cosn 2 2 sinn 2 cosn 2 2 sinn 2 sinn 2 sinn 1 2 s1n s1n2 s1nn SinwQ Taking real parts we get the formula 9 2 1 t9 2 11003910032911003n9 W s1n 2 We can also write Ema2 me2 62n102 7 Maw2 cosn 2 sinn 1 2 2 22 Ema2w 7 67202 7 6202 7 672n1292 4239 6202 7 67202 62n120 7 672n1202 4239 4239 sin 2 sinn 12 Tf39 So we can write 1 12 sin n 1cos cos2 cosn i W 5 POWERS AND ROOTS By DeMoivre7s formula if z rem rcos 9 2sin 9 then 2 Name r cos n19 2 sin n19 Now lets say we want to nd complex numbers C such that C 2 lf 2 rem then by the above formula we have that C Walen satis es C 2 But recall that 62 is 27T2 periodic That is z 28099 for any k E Z Thus C Weiw mln will work We need only take k 01 n 7 1 because after that the roots will repeat A complex number 2 rem therefore has n nth roots Wyen7 n T62lt027rn7 39 39 39 7 w62027rn71n39 Eg the 4th roots of 1 are 127 71 and 72 or 606i39r26M and elm2 In general 1 is also called unity77 so the nth roots of 1 are called the nth roots of unity 21m 2k 11 512k cosi2sini 201m71 n 77 We denote by run emfn the principal nth mat of unity and all the nth roots of unity are given by 2 n 1 Laumwmmwn CLASS NOTES WEEK 1 9 51 Facts and examples Example 1 on p 36 Show that 1 7 run 7 w 7 A 7 wg l 0 This one is interesting to see visually take all the nth roots of unity on the complex unit circle and add them together their center of mass is clearly the origin To see this analytically7 just use the standard geometric series trick 1711 7wn7wg77w271 17w 0 But as long as tun 3A 1 which is true for any n gt 1 this implies the result Problem 117 p 37 Solve z 715 25 This could be pretty tricky if we just jumped into it algebraically Lets see if we can use roots of unity to solve Divide both sides by 25 to get 17125 1 That means 1712 emf 5 for k 0172374 That means the solution is 1 z k01234 Pretty quick Problem 5c7 p 37 Find all values of i14 Well7 i 6quotquot27 so its 4th roots are eiquot82kquot4 k 0 17 2 3 Problem 5d7 p 37 Find all values of 1 7 33913 Well7 1 7 3i 26 M37 so its 3rd roots are W6i77r972k7r339 Quadratic formula Example 3 on p 36 Let abc be complex constants7 1 31 0 Show that the solutions of the equation 122 7 b2 7 c 0 are given by the usual quadratic formula 2 7 7bixb2 746w 7 2a 7 where vb2 7 4ac represents one of the two values of b2 7 4ac12 The solution here is the same as in the real number case It involves completing the square A very clean way to do it is to multiply by 4a7 then add b2 7 4ac the discriminant to both sides This gives 41222 7 4abz 7 b2 b2 7 4ac7 or 2oz 7 b2 b2 7 4ac Taking square roots7 we7re done 52 Introduction to Induction Problem 17 p 37 asks us to prove DeMoivre7s formula by induc tion That is7 we show that cos 9 7 i sin9 cos n9 7 i sin n9 for n 1727374757 The principle of induction is this First7 show that the statement holds for the rst n here for n 1 this is a trivial case Then7 show that if the statement holds for n7 then it also holds for n 71 Then it must hold for n 17 237 The case n 1 is trivial To show how the proof is supposed to proceed in a concrete way7 let7s show that the formula holds for n 2 Indeed7 we have cos 9 7 i sin92 cos 9 7 isin 9 cos 9 7 i sin 9 cos2 9 7 sin2 9 7 icos 9 sin9 7 sin 9 cos 9 cos 29 7 isin 29

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