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Complex Analysis I

by: Donny Graham

Complex Analysis I MTH 829

Donny Graham
GPA 3.57


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This 13 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 829 at Michigan State University taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/207310/mth-829-michigan-state-university in Mathematics (M) at Michigan State University.


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Date Created: 09/19/15
Math 829 Spring 2004 NOTES ON DIFFERENTIATION February 3 2004 1 The Chain Rule This is the following famous result 11 Theorem Suppose U and V are open sets with f and g complep Ualued func tions de ned on U and V respectively where fU C V Suppose that 20 E U so that f20 E V ff is samplers dz epehtz39able at 20 and g is dz epehtz39able at f20 then 9 o f U 7 C is dz epehtz39able at 20 and g o f 20 g f20f 20 PROOF Let we f20 Our hypotheses are that f2 e f20 f 202 e 20 122 where 522 7 0 as 2 e 20 1 and 900 7 9wo 9 wow 7 we sltwgt where 7 0 as w 7 we lt2 Our goal is to show that if h g o f then lT2l hz 7 h20 gf20f202 7 20 Tz where 7 0 as 2 7 20 3 l2 7 20 To this end substitute w fz into 2 legal because fU C V to get 112 h20 9 fzolf2 7 fzol Sf27 and then substitute the result of 1 into this equation to get hZ h20 9 fzof zo2 20 9 f20R2 502 4 The second term on the right side of 4 is olz 7 20 as 2 7 20 so all we have to do is show that the same is true for the third term on the right For this let 8w Swlw 7 wol so that Sw 8wlw 7 wol where 8w 7 0 as w 7 we Then substituting w fz and using 1 one last time 502 8f2lf2 fzol 8f2lf zol 2 Zol 8f2R27 where on the right hand side of the equation the rst term is olz 7 20 because fz 7 f20 as 2 7 20 remember differentiability at a point implies continuity there and the second term is olz 7 20 because Rz has this property and 8 is bounded Thus Sfz is olz 7 20 and z 7 20 and the proof of the chain rule is complete D Math 829 Spring 2004 12 Exercise Suppose G is an open subset of the plane ab is a nite closed real interval 7 ab a G is real differentiable at a point to 6 ab and f G a C is differentiable at yt0 Show that f o y is real differentiable at to and that f OWW f 7to7 to 2 Suf cient condition for differentiability We begin with a concrete situation Suppose G is an open subset of R2 and u G a R is a real valued function de ned on G Let 20 zoy0 be a point of G 21 Theorem If the rst partial derivatives ofu exist at every point ofG arid are continuous at 20 theri u is di ereritiable at 20 PROOF From our discussion in class it is enough to show that if Rh d f we h 7 u20 7 WW11 Wed12 for h h1h2 E R2 with lhl small enough that 20 h E G then lRhl olhl as lhl a 0 The rst step is to write u20 h UZo UWO 11790 hz 07y0 h2l UWOWO hz U07yol7 and then apply the one variable Mean Value Theorem of differential calculus to each of the square bracketed terms on the right With this you see that the right hand side of the equation above is UmWo 1790 h2h1 uy07yo t2h27 where t1 lies between 0 and hl and t2 lies between 0 and hg Thus 21 z0t1 y0h2 and 22 xmyo t2 both a 20 as h a 0 and PM mm 7 Umzolh1 lit22 7 Uyzolh27 5 for all h suf ciently small Let 81h and 82h denote respectively the two terms in square brackets on the right hand side of 5 these terms are functions of h because the points 21 and 22 depend only on h The continuity of um and uy at 20 used here for the rst and only time guarantees that that both 81h and 82h a 0 as h a 0 Thus the same is true of elthgt mamam l82hl Now an easy estimate starting with 5 yields lRhl S hlhl7 which shows that u is differentiable at 20 because as we just noted 8h a 0 as h a 0 D Math 829 Spring 2004 22 Exercise De ne differentiable77 for real valued functions de ned on open sub sets of R State and prove a suf cient condition for differentiability of such functions that generalizes Theorem 21 3 The cosmic truth about differentiation Suppose U is an open subset of R and that f U a R 31 De nition We say that f is di erentiable at a point p0 of U if there is a linear transformation T R a R such that fp0 h fp0 Th Rh Vh E R suf ciently close to pg7 6 where lRhl olhl as lhl a 0 in R 32 Notation If f is differentiable at p0 then the linear transformation T in the above de nition is called the derivative of f at p0 We will write T Df007 preferring almost always to reserve the prime77 notation for the complex derivative 33 Exercises You should go back to the de nition of differentiability for functions R2 a R and identify this linear transformation Do the same for functions from an open subset of R to R Do the same for differentiable functions from intervals of the real line to R2 or more generally R the so called vector valued functions Finally7 how do you t the de nition of complex differentiability77 into this linear transformation context 34 Exercise ff is di erentiable at p0 then f is continuous at p0 35 The matrix of Dfp0 Let 1627 en be the standard unit vector basis for R ie7 ej is the vector with 1 in the j th position and zeros elsewhere Then upon xingj and substituting h tej into 6 and lettingt a 07 an argument entirely similar to the one we used in class for the case n 2771 1 shows that Tej 6f 00 lt6f1100 100 7 57 57 quot 7 ax where f U a R is the j th coordinate function of f f f17f1m 7fm Thus the metrics of Dfp0 with respect to the standard bases in Rn and Rm respec tively is the one whose j th column is 37Cp0 written as a column vector7 rather than as the usual row vector Let7s call this matrix Dfp0 Thus Dfp0 33 190Z ZF1 The image Dfp0h of a vector h E R is the m dimensional vector found from the equation lDf100hl lDf100llhl7 3 Math 829 Spring 2004 where square brackets around a vector denote the corresponding column matrix ie the transpose of the original row vector 36 The Chain Rule revisited Now suppose that in addition to the setup above V is an open subset of R contained in fU and g V a RF is differentiable at fp0 Then 9 o f U a RF is differentiable at p0 and Dlt9 0 f100 D9f100Df100 where the product between the derivatives on the right is the product of linear trans formations ie their composition 37 Exercise Adopt the argument of 1 to prove this version of the Chain Rule Suggestion If you make a suitable de nition of the norm lTl of a linear transforma tion say lTl maxszl z E R 1 then the maximum in question exists and is nite because the map z a lTpl is a continuous real valued function on the compact unit sphere of R and you can easily prove that lTpl S Va 6 R 7 From this you should be able to write down a proof of the chain rule that is almost word for word the same as the one in 1 If you want a more concretely de ned norm for linear transformations you can take lTl to be the square root of sum of the squares of the entries of T a quantity that is in general larger than the previously de ned norm For this one the Cauchy Schwarz inequality gives 38 Corollary Suppose that o f U a R is di erentiable at each point of an open set U C R o I is an open interval of the real line and o y I a R is a di erentiable function with yI C U Suppose to E I and y t0 31 0 Then Df yt0 y t0 is a vector tangent to the image curve f 07 at f yt0 PROOF This is just a restatement of the Chain Rule D Corollary 38 says roughly that the linear map Dfp0 takes each vector tangent to a given curve through p0 into a vector tangent to the image curve at fp0 In this context when we assert that Dfp0 is a linear transformation from R to R we should actually think of R as the space of vectors tangent to all possible differentiable curves through p0 and R as the corresponding tangent space77 at fp0 This point of view will show up again in the next section 4 Math 829 Spring 2004 39 Exercise Suppose G is an open subset of R pg 6 G and u G a R is a real valued function de ned on G j 12 m Then f uh um G a R and every Rm valued function on G has this form State and prove a suf cient condition for differentiability of f at p0 that generalizes Theorem 21 to this situation Suggestion The problem quickly reduces to Exercise 22 4 Complex vs real differentiation In this section G will always be an open subset of R2 20 a point of G and f a function G a R2 This means that there are real valued functions u and 1 from G to R2 such that 7 u 2 2 fzivz R VZEG Suppose f is real differentiable at 20 with derivative Df20 at 20 Recall that this derivative is a linear transformation on R2 that satis es 6 above We know from our discussion ofthis real derivative that the rst partial derivatives of u and 1 with respect to r and y exist at 20 and that the matrix of Df20 is a a grim l loo where the notation on the right hand side means that all partial derivatives are eval uated at 20 Moreover it follows from 6 and the de nition of sampler differentiation that f is sampler differentiable at 20 precisely when it is real differentiable and the linear transformation Df20 can be effected by complex multiplication by a77 for some complex number a Now it was an exercise suggested in class part of which is rephrased in fancy algebraic language in Exercise 3 of page 2 of our text to show that the linear transformation induced on R2 by a 2 gtlt 2 real matrix Am is given by complex multiplication on R2 iff Oz 6 and y 76 in which case the transformation may be written 2 a a2 where a 04 2396 Putting these observations together we see that the function f initially assumed real differentiable at 20 is sampler dl erentz39able at that point iff the CaushyiRlemann equations 3U 31 an 311 3r 7 3y 3y 7 3r hold at that point Now the proof given in class that complex differentiability at 20 implies that the CauchyiRiemann equations hold there actually shows as a byprod uct that f u 2391 is real differentiable there with real derivative equal to the map 2 a f 20z the real linear transformation of R2 given by multiplication by the complex number f 20 Thus our discussion of differentiation has actually shown 5 Math 829 Spring 2004 Theorem Suppose G is an open subset of R2 with 2e a point of G Then 7 i 2 fzi G R is complew di erentiable i f is real di erentiable and the functions it and i satisfy the CauchyiRiemann equations at 2e 5 Differentiation and conformality Suppose G is open in R2 and f G a R2 is continuous on G and real differentiable at 2e 6 G Let we fze Suppose further that y ab a G is a curve in G that passes through 2e say yte 2e for some not necessarily unique te E a b and that y is differentiable at te with y te 31 0 we are thinking of y te as a two dimensional column vector hereiits coordinates are the t derivatives of the coordinate functions of 7 evaluated at te Suppose y te 31 0 Then y te is a tangent vector to y at 2e yte In this case lets say that y is regular at te Finally suppose the linear transformation f ze is nonsingular so that it takes non zero vectors to non zero vectors Then the Chain Rule tells us that f o y is real differentiable at te and f 0 WW f Vto to f 20V to 8 Thus the nonsingular linear transformation f ze acting on R2 takes the vector y te which is tangent to y at 2e to a vector tangent to the curve fo y at we fze In other words omitting much precision Di erentiable functions take smooth curves to smooth curves and their derivatives take tangent vectors to tangent uectors Now if f is complew differentiable at 2e then we think of f ze as a complex number which we7ll assume is non zero its an easy exercise to show that in this case this is equivalent to nonsingularity ofthe real derivative of f at 2e Thus f ze pew for some p gt 0 and o E R polar representation of a complex number and similarly y te rem Thus the image of the tangent vector y te is fZoWtol Wemwh which says that vectors tangent to y at 2e get rotated by f ze through an angle o arg f ze Thus two differentiable curves through 2e get taken to curves through we whose respective tangent vectors get rotated through the angle o arg f ze and because of this the angle between these image tangent vectors ie the angle between the image curves at we is the same as the angle between the original tangent vectors at 2e This preservation of angles is called conformality at 2e So far we have proved IfG is an open subset of R2 and f G a R2 is complew di erentiable at 2e 6 G with the linear transformation f ze nonsingular ie with the complex number f ze nonzero then f is conformal at 2e 6 Math 829 Spring 2004 What about the converse Suppose f is real differentiable at 20 and conformal there ls f complex differentiable at 20 with non zero complex derivative The answer is yes To see this view f 20 not as a complex number but as a linear transformation on R2 Let A be its matrix with respect to the standard basis of R2 Thus the columns of A are the images under f 20 of the standard unit vectors e1 eg of R2 ie Ae1 is the rst column of A and A62 the second Because f is conformal at 20 its derivative at 20 preserves the angles between vectors this is in fact the de nition of conformality so the columns of A are orthogonal to each other In addition they have the some length To see this observe that e1 7 e2 and el e2 are orthogonal hence so are their images under f 20 ie under multiplication by the matrix A Thus denoting dot product77 in R2 by 77 0 A61 A62 39 A61 7 A62 A61 39 A61 7 2A61 39 A62 7 A62 39 A62 HA61H2 7 HA6le where the last line follows from the orthogonality of Ae1 and A62 Thus A is a two by two matrix whose columns are orthogonal and have the same length I leave it as an exercise to you to show this means that for some real numbers a and b not both zero AZ ab or AZ fa 9 with the rst choice being the correct one because the linear transformation repre sented by A not only preserves the magnitudes of angles it also preserves their senses hence the image of e2 must lie 90 degrees counterclockwise from that of e1 Now if f u ii is the decomposition of f into real and imaginary parts then we know from the real differentiability assumed for f that u and i are real differentiable at 20 and so have rst partial derivatives at that point Our identi cation of the elements of the matrix of the real derivative of f at 20 with these partial derivatives along with the form discovered above forced on this matrix by the conformality of f at 20 shows that at 20 um a oy and 111 b iuy ie u and i obey the Cauchy Riemann equations at 20 and therefore f is complex differentiable at 20 To see that f 20 31 0 that in proving the equivalence of Cauchy Riemann and complex differentiability we also establish the fact that if f is complex differentiable at 20 then f 20 the complex number is equal to um20 iuy20 this comes from taking the limit of the difference quotient for f along the horizontal line through 20 That is f 20 viewed as a complex number is just a ib where a and b are the real numbers that show up in our characterization 9 of the matrix A 7 Math 829 Spring 2004 Conformality demands that A be nonsingular hence at least one of a b is nonzero ie f20 31 0 D Thus we have proved 51 Theorem Suppose G is an open subset of C and f u in G a C is real di erentiable at a point 20 E G Then the following are equivalent a f is complep di erentiable at 20 and f 20 31 0 b f is conformal at 20 In addition these arguments prove most of 52 Theorem Suppose G is an open subset of C and f u in G a C is real di erentiable at a point 20 E G Then the following are equivalent a f is complep di erentiable at 20 b u and i satisfy the Cauchy Riemann equations at 20 e The linear transformation f 20 can be identi ed on R2 complep multiplication on C by quotthe compleco number f 20 d The linear transformation f 20 on R2 is linear when viewed as a mapping of Sketch of proof We have already proved the equivalence of a and The equiva lence of these with c follows from the fact that in proving the equivalence of Cauchy Riemann and complex differentiability we also establish the fact that if f is complex differentiable at 20 then f 20 the complex number is equal to um20 iuy20 this comes from taking the limit of the difference quotient for f along the horizontal line through 20 In other words in the matrix representation 9 the rst column is the vector representation of the complex number f 20 and the matrix product A2 with a column vector 2 Lle yields the vector representation of the complex number f 20z The equivalence of c and d is just a restatement of the fact that any linear map C a C is multiplication by a xed complex number77 hint look at T1 D Math 829 Spring 2004 6 Conformality of the Stereographic Projection We apply the ideas of the previous sections to prove that the stereographic projection is conformal ie that differentiable curves in the plane that meet at a point 2 get projected to curves on 52 that meet at 2 and make the same angle with the same sense there 61 Notation For this discussion we denote 0 Points of R3 by i7 and those of R2 by my 0 The North Pole 001 of 2 by N o The inner product ie dot product of two vectors 1 and w in the same Eu clidean space by lt1 wgt 62 The stereographic extension77 Consider the natural extension 0 of the stereographic projection to R3 1 de ned by 6 77 7 7 10 U 77C 141 lt gt By the suf cient condition of Exercise 39 o is differentiable on M dEf R3C 1 and by our work in class the matrix of the derivative of o 0102 at a point p i7 E M is obtained by placing the vector partial derivative of o with respect to each coordinate down the respective columns of a two by three matrix p p p 1 1 0 U as an ac 1 D W gem gimp 35ml 1ltl0 1 From our discussion of the chain rule the problem is to show that for every point p i7 E SZN the linear transformation Dop preserves angles between vectors tangent to 2 at pl In other words if i and w are three dimensional vectors tangent to 2 at p analytically their inner products with p are both zero then the angle from o to w is the same as the angle from Dopo to Dopw 63 Analytic statement of the problem Its pretty clear that the stereographic projection preserves the sense of angles so we will concentrate on preservation of magnitudes of angles Here is the analytic expression of what needs to be done Suppose p E SZN and ow E R3 with lt10 Ugt 0 Wd lt10 10gt 07 11 ie o andw are orthogonal to the line from the origin to p and hence tangent to 2 at p Let A Dop the matrico of Dfp with respect to the standard basis We desire to show that A A lt v wgt ltv wgt 12 WA mm m lwl 9 Math 829 Spring 2004 The quantities that show up on the left and right hand sides of this last equation are you will recall the cosmes of the angles between the respective pairs of vectorsl 64 Exerciseireduction of problem Show that in order to prove 12 you need only show that there is a positive constant c such that Al Aw 01 w 13 for all 1110 6 R3 satisfying 11 In order to prove 13 we observe that Al Aw AVM w where A is the linear transformation R2 a R3 whose matrix is the transpose of the matrix of A This is just a matrix calculation based on the fact that x y matrix product where z and y are any vectors in the same Euclidean space and are their respective column vectors and the superscript T77 denotes matrix transpose Thus HUMAN AwlTlAvllAllwlTlAllvl lwlTlAlTlAllvl lwlTlAiAllvl w AVMI from which the desired result follows by the symmetry of the inner product it is unchanged if the order of its entries is reversed Note that this calculation works as well for any m gtlt 71 matrix and any pair of vectors in R 65 Computations with AA For p 77C E SZN let7s write 77 7 d 7 14 z 7 C an y 1 7 C lt gt We use z and y simply as notational conveniences here but nevertheless note that z W is the stereographic image in the complex plane of the point p 77C of 82 From the work of 62 the matrix of A dEf DUp can be written AF HTZl whereupon 1 0 AM A A c 0 1 y 2 2 1Remember that the inner product on the righthand side of the equation is the one for R3 and the one on the left is that of R21 10 Math 829 Spring 2004 1 1 Suppose for the moment that 1 111112113 and w 111171027103 are any vectors in R3 Then using the fact that ltAv Awgt ltAAv wgt wlTlAVi 1 we obtain after a little calculation where c c1ltAvAwgt mm wgt lwlTlAAllvl 11 96119101 12 yvsw2 95111 1112 952 yZWSlws vlwl 1ng 3103 zwl yw2v3 mm yvz 2 12 7 1v3w3 lt1 wgt A where A 101 ywz is lmi 1112 2 12 1 3lw3 We claim that ifi and w are tangent to 2 at p then A 07 which will nish our proof For this7 go back to equation 14 describing z and y in terms of the coordinates of p7 and note that 2C 2 2 z 7 1 7 y 1 7 lt7 hence 1 0A 5101 WWW 511 7712 2C03w3 5101 77102 Cws s 5111 7702 9103 lt107 W113 lt107 gt103 lf 1 and w are tangent to 2 at p7 then the inner products in the last line are both zero7 hence A 0 since p 31 N i C 31 1 This completes the proof that the stereographic projection is conformal D 11 Math 829 Spring 2004 7 The Mercator Projection This section borrows heavily from the beautiful book by Eli Maor Trigonometm39c Delights Princeton University Press 1998 especially Chapters 13 and 14 71 Mercator conformality and calculus In order to make a map ofthe earth one has to address the problem of representing the sphere 2 with some accuracy on a at plane The Flemish map maker Gerardus Mercator attacked the problem of making a map that would represent SZKl lt 1 7 8 the unit sphere of R3 with equal spherical caps removed around the north and south poles on a rectangle 7 7T gtlt lib h where 8 is some small positive number and h is positive Mercator wanted his map to have the following properties a The circles of latitude should be represented by horizontal lines of length 2h with the equator latitude zero degrees the horizontal axis of symmetry b Equally spaced circles of longitude great circles on 52 through the north and south poles should be represented by equally spaced vertical lines c The correspondence between points of 2 and points of Mercator7s map should be conformal Conformality is the most important propertyiit guarantees that at least in principal a traveler wishing to go from point A on the globe to point B need only draw a line between the corresponding points on Mercator7s map measure the angle between this line and the vertical axis true north and using a compass travel along a path that always makes the desired constant compass heading Such a map cannot of course accurately represent distances Indeed the circles of latitude on 52 get smaller as their centers approach the poles and the Mercator representation must stretch distances along these circles to make the corresponding horizontal lines the same length This is why Greenland for example looks immense on a Mercator projection when in reality it is not In order to preserve conformality in the face of such horizontal stretching Mercator had to correspondingly distort the distances between the lines of latitude for example the vertical distance on Mercator7s map between the lines representing latitude 15 and 30 will be larger than that between the equator and the line representing latitude 15 Mercator7s great triumph was to gure out how to accomplish the vertical stretching that insures conformality He published his map of the world in 1569 but unfortunately never explained his method an omission that contributed to a certain skepticism about the value of his accomplishment In 1599 Edward Wright lifted the veil of mystery from Mercator7s method publish ing an accurate account of its underlying mathematics Nowadays the mathematical foundation of Mercator7s feat reduces to an exercise in freshman calculus as is de scribed succinctly in Chapter 13 of the above mentioned book by Maor There it7s shown how in order for spherical rectangles to get mapped to similar plane rectan gles the height y of the line representing the circle of latitude A on the sphere must 12 Math 829 Spring 2004 obey the differential equation dy secAdA 15 in other words A y sec tdt 16 0 Now Wright did not phrase his solution in terms of calculusiindeed his work ap peared long before Newton and Leibnitz fully developed the subject What Wright did was solve the discrete analogue of 15 in increments of one degree Nowadays we prove in freshman calculus at least we used to before Calculus Reform that A A sectdtln tan 7I 17 0 2 4 but note that it wasnt until 1614 fteen years after Wright7s treatise on Mercator7s projection that Napier published his invention of logarithms In 1645 Henry Bond based on both the work of Wright and recently published tables of logarithms conjectured 17 and this became one ofthe outstanding mathe matical problems of the latter half of the seventeenth century James Gregory proved 17 in 1668 but his proof was so complicated that it was viewed as suspicious at best Two years later lsaac Barrow Newton7s predecessor at Cambridge gave a com prehensible proofiessentially the same one you nd in the calculus books of today For this Barrow invented the technique of partial fraction decompositions which he applied to the evaluation of many other integrals 72 The Mercator Projection and the complex logarithm The work we have done in these notes on conformality ofthe stereographic projection the corresponding conformality of holomorphic functions done in class and the holomorphicness of the Principal Branch of the logarithm function result in a quick solution of Mercator7s problem a First note that the stereographic projection maps the spherical region SS dEf SZKl lt 1 7 8 onto the annular open subset of C described by AR26C1RltlzlltR where R x 2 7 8 8 gt 1 This is a little exercise in right triangles along with the formulas describing the stereographic projection and one of the re ection results you obtained in the problem set about that projection I leave it to you The actual relationship between R and 8 is not important what is important is that the inner radius of the annulus is the reciprocal of the outer radius A C7 V Recall that under the stereographic projection the circles of longitude on the sphere go to rays in the plane and the circles of latitude go to circles in the plane centered at the origin 13


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