j=1 dz Res(f, z j ) = 1 d k 1 dz k 1 (z c)k f(z) Res(f, c) = lim z c (k 1)! Res g, c = f(c) g (c)

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1 Problem. Compute the itegrals C r d for Z, where C r = ad r >. Recall that C r has the couter-clockwise orietatio. Solutio: We will use the idue Theorem to solve this oe. We could istead use other (perhaps more elemetary methods) but this is a good warm-up. Below is a brief review of the theorems which we will use. Theorem (idue Theorem). Let Ω C be a ope set, ad let,, m Ω. Suppose that is a ull-homotopic closed loop i Ω, ad does ot pass through ay of the poits,, m. Let f be a holomorphic fuctio i Ω {,, m }. The we have m f = 2πi Id(, j )(f, j ) j= Recall that Id(, j ) = d 2πi c (f, j ) = f()d 2πi D ɛ( j) We will use the properties proved i class (a) If a is the st coefficiet i the Lauret series of f aroud c, the (f, c) = a (b) If Ord(f, c) (removable sigularity) the (f, c) = (c) If Ord(f, c) = k (a k-pole), the (f, c) = lim c (k )! d k d k ( c)k f() (d) If Ord(f, c) ad Ord(g, c) =, the ( ) f g, c = f(c) g (c) (e) (, c) is C-liear i O(D x r (c)). We chaged a few of these from their statemets i the otes. Most importatly, we chaged property (c) ad (d). The ew property (c) is easy to prove, ad it is left to the reader. Proof of the ew property (d). The case Ord(f, c) = is proved i the otes. Whe Ord(f, c) >, the f() = ( c) f() for some holomorphic f. The we have f() g() = f() g()/( c) it follows that g()/( c) has order at c, ad so the above expressio has order or greater, ad thus it s residue is. However, we also have f(c) = because f has order greater tha at c. It follows that the

2 formula remais true. Now to solve the problem usig the residue theorem, we coclude d = 2πi (, ) Sice is holomorphic i Dr x, ad clearly has idex aroud. It is clear that Ord(, ) is, ad so by property (b) above, we coclude that d = if Next, by applyig property (d) with f() = ad g() =, we have (, ) = = ad so Now by property (c), we coclude that = 2πi. Ord(, ) = lim ( )! d d ( ) = for >, (we remark that we could have used this to fid (, ), but we are practicig usig all the properties) ad so we obtai Problem 2. Evaluate the itegrals = for >. + x m dx + x ad π a + si 2 θ where m, are itegers satisfyig m 2 ad a > is real. Solutio: For the first oe we will itegrate alog the cotour show below (t) = t, t [, R] 2 (t) = t exp (2πi/), t [, R] 3 (t) = R exp (it), t [, 2πi/] It is clear that the oly ero of + iside the iterior of our cotours is at = exp (πi/), ad thus by the idue theorem, we ca coclude that +3 2 m ( ) m d = 2πi + +, exp(πi/) 2

3 2 3 = exp (πi/) Figure : Cotour used i the evaluatio of the first itegral To fid the residue we expad + aroud = u + exp (πi/), where we obtai a expasio begiig with the term exp (πi/)u. This proves that Ord( +, exp (πi/)) =, ad so we ca use property (c) of residues above to coclude that ( ) m +, exp(πi/) m m + = lim = = m+ = ( exp(πi/) exp iπ m + ) Now we ote that o 2 we have m exp (2πi(m/))t m, + = + t ad d exp (2πi/)dt, ad so we have Combiig these, we obtai 2 m ( + = exp 2πi m + ) m + ( ( exp 2πi m + )) m + 3 d + m ( d = 2πi + exp iπ m + ) Whe we take the limit R, we ca coclude that the itegral over 3 vaishes, sice the itegrad decays at least like 2, ad the legth of the curves grows oly like, so the itegral decays like. We also ote that the itegral we wish to evaluate is f()d as R. Dividig by 2i exp (iπ(m + )/), we obtai We ote that this implies ( ( exp πi m + ) ( exp πi m + )) 2i lim R R t m + t dt = π ( si π m + ) t m + t dt = π = t m + t dt = π ( csc π m + ) For the secod itegral, we ote by symmetry of si that π a + si 2 θ = 2π 2 a + si 2 θ 3

4 Now let = exp (iθ), whereby we obtai d = i, ad si θ = / 2i = si 2 θ = We also ote that whe we itegrate θ from to 2π, we are itegratig the variable over the uit disk. We obtai 2π 2 a + si 2 θ = 2i D d 4a = 2i D d 4 (4a + 2) 2 + Now to fid the residues, we must fid the eroes of 4 (4a + 2) 2 +. Completig the square, 4 (4a + 2) 2 + = ( 2 (2a + )) 2 4a 2 4a Which implies that the set of eroes is the four roots r ±,± = ± 2a + ± 2 a(a + ) Sice a >, we ca coclude that r ±,+ >, ad so it lies outside of D. Sice 2a + 2 a(a + ) > implies 4a 2 + 4a + 4a 2 + 4a a(a + ) ad a, we coclude that r ±, D. Fially we ote that 2a + 2 a(a + ) = implies that =, a cotradictio. This meas that 4 (4a + 2) 2 + has order at each of the roots, ad so we ca apply property (c) of residues to coclude ( ) 4 (4a + 2) 2 +, r ±, = 4(r ±, ) 2 4(2a + ) = 8 a(a + ) We coclude that the itegral is 2i D d 4 (4a + 2) 2 + = 2i 2πi ± 8 a(a + ) = π a(a + ) Problem 3. Calculate the residues of ta(π) ad cot(π) at their poles. Solutio: Usig the fact that ta π = si π/ cos π, we ca coclude that wheever ta π has a pole, we have cos π =. As we proved i assigmet 2, cos = implies that = π/2 + π for some Z. Sice cos (π/2 + π) = ± we coclude that cos has order at it s eroes. We ca therefore apply property (c) of residues ad coclude that Therefore ta has residue π ( ) si π cos π, r = si π si = π at each of it s poles. Applyig the same exact aalysis to cot, we obtai ( cos ) si, r = cos π cos = π Ad so cot has residue π at each of it s poles. 4

5 Problem 4. Show that the sum of the residues of a ratioal fuctio (together with the residue at ) is equal to ero. As part of this exercise you eed to itroduce a atural defiitio for the residue at. Solutio: Defiitio. The residue at ifiity of a fuctio f OD r,, where D r, is a aulus with a ifiite outer radius, is defied to be (f, ) = f()d, A few remarks:. Sice we require f to be holomorphic o D r,, we ca ot have ay sigularities i D r,. 2. By homotopy of, ad D r,, this defiitio does ot deped o the r chose. 3. By our choice of orietatio of,r [show below] we coclude that, =. Figure 2: Orietatio of R Let f be a ratioal fuctio, the f = p/q, where p ad q are polyomials. Clearly f is meromorphic, ad it s isolated sigularities are the roots of the polyomial q. Sice q has fiitely may roots, f has fiitely may sigularities {,, }. Let D r be a disk so large that it cotais all of {,, }. The by the residue theorem, we ca coclude that f()d = 2πi (f, i ) () However, i this sum we are omittig the residue at ifiity. Sice all of the sigularities are i D r we ca coclude that f is holomorphic o D r,, ad so i= (f, ) = f()d = f()d (2) 2πi, 2πi D R 5

6 where we used =,. Addig () ad (2), we coclude (f, ) + (f, i ) = Problem 5. Let f be holomorphic ad bouded i a puctured eighbourhood of. Show that is holomorphic i a eighbourhood of. i= 2 f() for g() = for = Solutio: For cocreteess, let f be holomorphic o D r (). To prove that g() is holomorphic o D r (), we will simply prove that it is differetiable at =. Sice we already kow that it is differetiable everywhere o D r (), we will have solved the problem. To prove that g is differetiable at, cosider the differece quotiet: g() g() 2 f() lim = lim = lim f() Now sice f is bouded, we ca coclude that this limit goes to ero, ad thus g is differetiable at. It follows that g is holomorphic everywhere. Oe cosequece of the proof is that g has g() = ad g () =, this allows us to coclude that g() = 2 h() for some holomorphic h. Notice that this result gives a alterate proof of the removable sigularity theorem: sice f() = g()/ 2 = h() everywhere i D r () we ca use the idetity theorem to coclude that f has a removable sigularity at. Problem 6. Let c C be a isolated essetial sigular poit of f. Prove that for ay give α R ad a arbitrary small r >, there exists D r (c) { c } such that Rf() = α. Solutio: We will use two theorems proved i the course otes: Theorem (Casorati-Weierstrass Theorem). Let f O(D x r (c)), ad let c be a essetial sigularity of f. The f(d x r (c)) is dese i C. Theorem (Preservatio of Domais). If Ω C is a coected ope set ad f O(Ω) is ocostat, the f(ω) is also a coected ope set Now suppose that there exists a α R such that there does ot exist a Dr x (c) such that Rf() = α. The the lie l = { C R = α } is disjoit from f(dr x (c)). Sice f(dr x (c)) is dese, we ca coclude that there exist poits of f(dr x (c)) o either side of the lie l. Let C α ad C +α be the two ope halves of the complex plae C α = { C R < α } ad C +α = { C R > α } 6

7 It is clear that these disjoit ope sets cover f(dr x (c)), ad both C α f(dr x (c)) ad C +α f(dr x (c)) are o-empty. Therefore f(dr x (c)) ca ot be coected. However, by the Preservatio of Domais theorem, ad the fact that Dr x (c) is coected, we must have f(dr x (c)) coected. This is a cotradictio, ad so we must coclude the existece of some Dr x (c) such that Rf() = α. Problem 7. Let : [a, b] A be a piecewise differetiable curve, where A C is a ope set, ad let g : A Ω C be a coutiuous fuctio of two complex variables, where Ω C is also a ope set. Assume that for ay fixed w A, the fuctio g(w, ) is holomorphic i Ω. The prove that f() = g(w, )dw is holomorphic i Ω. Solutio: Let c Ω. Sice g(w, ) is holomorphic i Ω for each w, we ca coclude that g(w, ) = g(w, c) + h(w, )( c) For some h(w, ) cotious at c (i the secod argumet, holdig the first oe fixed). Furthermore, sice h(w, ) = g(w, ) g(w, c) ( c) we ca coclude that h is also cotiuous i the first argumet (holdig the secod oe fixed). However, if a fuctio is cotiuous i both argumets seperately, we ca coclude that it is cotiuous. The we ca coclude that f() = g(w, c) + h(w, )( c) dw = f(c) + ( c) h(w, ) (3) It is clear that the itegral h(w, )dw is well defied by cotiuity of h. Furthermore, it is a basic fact of itegratio that h(w, )dw is a cotiuous fuctio of. Therefore the equatio (3) implies that f is differetiable at c. Sice c was arbitary, we coclude that f is holomorphic o Ω. Note: We could have also used Morera s theorem i the maer we have used to prove the Weierstrass covergece theorem (ad the Schwar reflectio priciple). proof: h(w + δ, + ɛ) h(w, ) = h(w + δ, + ɛ) h(w, + ɛ) + h(w, + ɛ) h(w, ) 7

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