SOLUTIONS TO HOMEWORK ASSIGNMENT # 7

1 SOLUTIONS TO HOMEWORK ASSIGNMENT # 71. Determine the nature of all singularities of the following functionsf(z).(a)f(z) = cos 1/z.(b)f(z) =1z2sinz.(c)f(z) =zez2 :(a)z= 0 is the only singularity. It is an essential singularity since the Laurent seriesexpansion aboutz= 0,cos 1/z= 1 12!z2+14!z4+ ,has infinitely many negative powers ofz.(b) The singularities arez= 0 andz=n , n= 1, 2, ..The singularity atz= 0is a pole of order 3 sincez= 0 is a zero of order 3 follows easily fromthe Maclaurin series aboutz= 0 :z2sinz=z3 13!z5+15!z7+ = n=0( 1)n1(2n+ 1)!z2n+ singularitiesz=n , n= 1, 2, .. ,are simple poles since they are simple zerosofz2sinz.(c)z= 0 is a simple pole sincezez2 1=zz2+z4/2! +z6/3! + =1z+z3/2! +z5/3! + =1zg(z)whereg(z) is analytic atz= 0 andg(0)6= factg(0) = 1,although what simportant is just thatg(0)6= other singularities are the non-zero SOLUTIONS ofez2= 1,that isz= 2n ,wherenis a non-zero integer.

2 They are all simple poles sinceddz(ez2 1)|z= 2n = 2 2n e2n = 2 2n 6= Evaluate the following integrals. In each case the contour is positively oriented.(a) |z|=R zndz,wherenis an integer.(b) |z|= (c) |z 1|= :(a) Make the substitutionz=Re .Thendz=R e d and so |z|=R zndz= =2 =0 Rn+1e( n+1) d = Rn+1 =2 =0e( n+1) d ={2 R2n= 10n6= 1It is obvious =2 =0e( n+1) d = 2 ifn= 1 then the Fundamental Theoremof Calculus gives =2 =0e( n+1) d =e( n+1) n+ 1 =2 =0= 0 The point to this question is that the functionf(z) = zis not analytic, for if it werethe Cauchy Integral Theorem would tell us that |z|=R zndz= 0 forn 0.(b) This is a straight forward application of the Cauchy Residue Theorem: |z|=3cotzdz= 2 Residue(cotz, z= 0) = 2 zcoszsinz z=0= 2 .The singularities of cotz=coszsinzarez=n , n= 0, 1, 2.}

3 They are all simplepoles, but only the singularity atz= 0 is inside the cirlce|z|= 3.(c) The singularities of1zsinzinside the circle|z 1|= 4 arez= 0 andz= .Thesingularity atz= 0 is a pole of order 2 since the Laurent series atz= 0 is1zsinz=1z2(1 z2/3! +z4/5! + )=1z2+16+ Here we have used the geometric series:1zsinz=1z(z z3/3! +z5/5! + )=1z2(1 z2/3! +z4/5! + )=1z2(1 (z2/3! z4/5! + ))=1z2(1 + (z2/3! z4/5! + ) + (z2/3! z4/5! + )2+ )=1z2+ 1/3! + higher powers ofzTherefore the residue atz= 0 is way to see this is that1zsinz=1z2g(z) whereg(z) =zsinzNow we could expandg(z) =z/sinzas a Taylor series aboutz= sinceg(z) is aneven function it follows that the Taylor series will have the forma0+a1z2+a4z4+ ,and therefore the residue atz= 0 is don t actually have to compute the singularity atz= is a simple pole and therefore the residue atz= isz zsinz z= = 1/.

4 Therefore |z 1|=41zsinzdz= 2 .3. Letf(z) be the power series n=0n2zn.(a) Find allzsuch that the power series converges.(b) Find a closed form expression forf(z).Solution:(a) By the ratio test the series converges for|z|<1 and diverges for|z|> seriesdiverges for|z|= 1 since the termsn2zndo not go to 0 asn if|z|= 1.(b) Consider the geometric series11 z= 1 +z+z2+z3+ Thenzddz(1 z) 1=z+ 2z2+ 3z3+ Do it one more time:z+ 22z2+ 32z3+ =zddz(zddz(1 z) 1)=zddz(z(1 z) 2) =z(1 +z)(1 z)34. Find allzsuch that the power series n= : By the ratio test we see that n=11n2znconverges for|z|<1 and diverges for|z|> also converges for|z|= 1 by comparison with the series n= Supposef(z) is analytic for|z| 1 and|f(z)| Mfor|z|= 1,whereMis someconstant. Show that|f(0)| Mand|f (0)| : This follows from a Cauchy Integral Formula and theM Linequality:f(0) =12 |z|=1f(z)zdz= |f(0)| 12 M2 =Mf (0) =12 |z|=1f(z)z2dz= |f (0)| 12 M2 =MExercise:What inequalities do you get for|f(n)(0)|?

5 6. Determine if there is a functionf(z) which is analytic in some open neighbourhood ofthe origin and which satisfies the following. If there is such a function find a closedform for it and state wheref(z) is analytic.(a)f(k)(0) =kfork 0.(b)f(k)(0) = (k!)2fork 0.(c)f(0) = andf(k)(0) = ( 1)k+12k(k 1)! fork : In all cases we consider the Maclaurin seriesf(z) = k=0fk(0)k!zk.(a)f(z) = k=0fk(0)k!zk= k=11(k 1)!zk= (z) is entire.(b) In this case we would havef(z) = k=0k!zk,which diverges for allz6= thereis no such function.(c)f(z) = + k=1( 1)k+12k(k 1)!k!zk= + k=1( 1)k+11k(2z)k= +Log(1 + 2z).This converges for|z|<1 Evaluate the following integrals. In each case the contour is positively oriented.(a) CR1z2+z+ 1dz,whereR >1 andCRis the real axis from RtoRtogetherwith the upper half of the circle|z|=R.(b) |z|=1z2e1/zsin(1/z) :(a) The singularities off(z) =1z2+z+ 1occur at the roots ofz2+z+ onlyroot inside the contourCRisz=e2 /3,and it is a simple pole.

6 Thus4 CR1z2+z+ 1dz= 2 Residue(1z2+z+ 1, z=e2 /3)= 2 z e2 /3z2+z+ 1 z=e2 /3= 2 12z+ 1 z=e2 /3=2 3(b) The only singularity ofz2e1/zsin(1/z) occurs atz= 0,and it is an essentialsingularity. Therefore the formula for computing the residue at a pole will not work,but we can still compute some of the coefficients in the Laurent series expansion aboutz= 0 :z2e1/zsin(1/z) =z2(1 +1z+12!z2+13!z3+ )(1z 13!z3+15!z5 + )=z2(1z+1z2+(12 16)1z3+ )=z+ 1 +13z+ = Residue(z2e1/zsin(1/z), z= 0) =13 Therefore |z|=1z2e1/zsin(1/z)dz=2 :Read about the Cauchy product in the Evaluate 0x2+ 1x4+ :Consider the integral CR1 +z21 +z4dz,whereR >0 andCRis the positively orientedcontour comprised of the segment of the real axis from RtoRand then the upperhalf of the circle|z|= R,C Rdenote the real axis portion and the circularportion resp.

7 Then limR C R1 +z21 +z4dz= 0 since the degree ofz4+ 1 is 2 more thanthe degree ofz2+ singularities are at the SOLUTIONS of the equationz4+ 1 = 0,that isz=e /4, z=e3 /4, z=e5 /4, z=e7 only singularities in the upper half plane arez=e /4, z=e3 /4,and they aresimple poles. It follows that x2+ 1x4+ 1dx= 2 (Residue(1 +z21 +z4, e /4)+Residue(1 +z21 +z4, e3 /4))= 2 ((z e /4)(1 +z2)1 +z4 z=e /4+(z e3 /4)(1 +z2)1 +z4 z=e3 /4)5= 2 (1 + 4e3 /4+1 4e /4)= 2((1 )e3 /4+ (1 + )e /4))= 2((1 )( 1 + 2)+ (1 + )(1 + 2))= 2 2( (1 )2+ (1 + )2)= 2 Therefore 0x2+ 1x4+ 1dx= :In this calculation we have used the fact that limR C RP(z)Q(z)dz= 0,whereP(z), Q(z) are polynomials such thatdegree(Q) degree(P) + page322. The basic reason for this is thatP(z)Q(z)behaves like 1/Rdon the arc, whered=degree(Q) degree(P); whereas the arc only has length theM Linequality guarantees that the integral goes to Evaluate 11 + sin2 d.

8 Solution:We make the substitutionz=e .Thendz= e d = zd ,ord =dz =e e 2 =z 1/z2 .Therefore 11 + sin2 d = |z|=11 z(1 +(z 1/z2 )2)dz= |z|=11 z(1 z2 2+1/z24)dz=4 |z|=11z(6 z2 1/z2)dz=4 |z|=1z6z2 z4 1dzThe singularities occur at SOLUTIONS ofz4 6z2+ 1 = 0,that isz= 3 2 them are simple poles, but onlyz= 3 2 2 are inside the circle|z|= compute the residues at these singularities:Residue(z6z2 z4 1, z= 3 2 2)=z(z 3 2 2)6z2 z4 1 z= 3 2 2= 3 2 2 4(3 2 2)3/2+ 12 3 2 2=1 4(3 2 2) + 12=18 2In a similar manner we calculate thatResidue(z6z2 z4 1, z= 3 2 2)=18 11 + sin2 d =4 2 (the sum of the residues) =4 2 14 2= 210. Show that 1(1 +x2)n+1dx= (2n)!22n(n!)2forn= 0,1,2, ..Solution: We use an argument similar to that used in question 8. In particular see theremark at the end of that question. The only singularity of1(1 +z2)n+1in the upperhalf plane is atz= ,and it is a pole of ordern+ 1(1 +x2)n+1dx= 2 Residue(1(1 +x2)n+1, z= )=2 n!

9 Dndzn(z )n+1(1 +z2)n+1 z= =2 n!dndzn(z+ ) n 1 z= =2 n!( n 1)( n 2) ( n n)(2 ) 2n 1=2 n!( 1)n(n+ 1)(n+ 2) (2n)(2 )2n+1= 22n(n+ 1)(n+ 2) (2n)n!= (2n)!22n(n!)27