Mathematics312(Fall2013) November25,2013 Prof. Michael Kozdron

Lecture #31: The Cauchy Residue Theorem

Recall that last class we showed that a function f(z)hasapoleoforderm at z0 if and only if g(z) f(z)= (z z )m − 0 for some function g(z)thatisanalyticinaneighbourhoodofz and has g(z ) =0.Wealso 0 0 ￿ derived a formula for Res(f; z0).

Theorem 31.1. If f(z) is analytic for 0 < z z0

m 1 m 1 1 d − m 1 d − m Res(f; z0)= m 1 (z z0) f(z) = lim m 1 (z z0) f(z). (m 1)! dz − − (m 1)! z z0 dz − − ￿z=z0 → − ￿ − ￿ In particular, if z0 is a simple pole, then ￿

Res(f; z0)=(z z0)f(z) =lim(z z0)f(z). − z z0 − ￿z=z0 → ￿ ￿ Example 31.2. Suppose that ￿ sin z f(z)= . (z2 1)2 − Determine the order of the pole at z0 =1. Solution. Observe that z2 1=(z 1)(z +1)andso − − sin z sin z sin z/(z +1)2 f(z)= = = . (z2 1)2 (z 1)2(z +1)2 (z 1)2 − − − Since sin z g(z)= (z +1)2 2 is analytic at 1 and g(1) = 2− sin(1) =0,weconcludethatz =1isapoleoforder2. ￿ 0

Example 31.3. Determine the residue at z0 =1of sin z f(z)= (z2 1)2 − and compute f(z)dz ￿C where C = z 1 =1/2 is the circle of radius 1/2centredat1orientedcounterclockwise. {| − | }

31–1 Solution. Since we can write (z 1)2f(z)=g(z)where − sin z g(z)= (z +1)2 is analytic at z =1withg(1) =0,theresidueoff(z)atz =1is 0 ￿ 0 1 d2 1 d Res(f;1)= − (z 1)2f(z) = (z 1)2f(z) (2 1)! dz2 1 − dz − − − ￿z=1 ￿z=1 ￿ d sin z ￿ ￿ = ￿ ￿ dz (z +1)2 ￿ ￿z=1 (z +1)2 cos￿z 2(z +1)sinz = ￿ − (￿z +1)4 ￿z=1 4cos1 4sin1 ￿ = − ￿ 16 ￿ cos 1 sin 1 = − . 4 Observe that if C = z 1 =1/2 oriented counterclockwise, then the only singularity of {| − | } f(z)insideC is at z0 =1.Therefore,

sin z (cos 1 sin 1)πi dz =2πi Res(f;1)= − . (z2 1)2 2 ￿C − It is worth pointing out that we could have also obtained this solution using the Cauchy Integral Formula; that is,

sin z g(z) cos 1 sin 1 (cos 1 sin 1)πi dz = dz =2πig￿(1) = 2πi − = − (z2 1)2 (z 1)2 · 4 2 ￿C − ￿C − as above.

Remark. Suppose that C is a closed contour oriented counterclockwise. If f(z)isanalytic inside and on C except for a single point z0 where f(z)hasapoleoforderm,thenboth the Cauchy Integral Formula and the residue formula will require exactly the same work, namely the calculation of the m 1derivativeof(z z )mf(z). − − 0 Recall that there are two other types of isolated singular points to consider, namely removable singularities and essential singularities. If the singularity is removable, then the residue is obviously 0. Unfortunately, there is no direct way to determine the residue associated with an essential singularity. The coefficient a 1 of the Laurent series must be determined explicitly. − In summary, suppose that f(z)isanalyticfor0< z z isolated singularity is removable, a pole, or essential. If it believed to be either removable or essential, then compute the Laurent series to determine Res(f; z0). If it is believed to be a pole, then attempt to compute Res(f; z0)usingTheorem30.6.

31–2 Theorem 31.4 (Cauchy Residue Theorem). Suppose that C is a closed contour oriented counterclockwise. If f(z) is analytic inside and on C except at a finite number of isolated singularities z1,z2,...,zn, then

n

f(z)dz =2πi Res(f; zj). C j=1 ￿ ￿ Proof. Observe that if C is a closed contour oriented counterclockwise, then integration over C can be continuously deformed to a union of integrations over C1,C2,...,Cn where Cj is a circle oriented counterclockwise encircling exactly one isolated singularity, namely zj,and not passing through any of the other isolated singular points. This yields

f(z)dz = f(z)dz + + f(z)dz. ··· ￿C ￿C1 ￿Cn Since

f(z)dz =2πi Res(f; zj), ￿Cj the proof is complete. Remark. Note that if the isolated singularities of f(z)insideC are all either removable or poles, then the Cauchy Integral Formula can also be used to compute

f(z)dz. ￿C If any of the isolated singularities are essential, then the Cauchy Integral Formula does not apply. Moreover, even when f(z)hasonlyremovablesingularitiesorpoles,theResidue Theorem is often much easier to use than the Cauchy Integral Formula. Example 31.5. Compute 3z3 +4z2 5z +1 − dz (z 2i)(z3 z) ￿C − − where C = z =3 is the circle of radius 3 centred at 0 oriented counterclockwise. {| | } Solution. Observe that 3z3 +4z2 5z +1 3z3 +4z2 5z +1 f(z)= − = − (z 2i)(z3 z) z(z 1)(z +1)(z 2i) − − − − has isolated singular points at z1 =0,z2 =1,z3 = 1, and z4 =2i.Moreover,eachisolated singularity is a simple pole, and so −

3z3 +4z2 5z +1 1 i Res(f;0)= − = = , (z 1)(z +1)(z 2i) 1 2i −2 ￿z=0 − − ￿ − ·− 3z3 +4z2 5z +1 3+4￿ 5+1 3 3(1 + 2i) Res(f;1)= − = −￿ = = , z(z +1)(z 2i) 1 2 (1 2i) 2(1 2i) 10 ￿z=1 − ￿ · · − − ￿ ￿ 31–3 3z3 +4z2 5z +1 3+4+5+1 7 7(2i 1) Res(f; 1) = − = − = = − , − z(z 1)(z 2i) 1 2 ( 1 2i) −2(1 + 2i) 10 ￿z= 1 − − ￿ − − ·− · − − 3z3 +4z2 5z +1￿ 3(2i)3 +4(2i)2 5(2i)+1 34 15i Res(f;2i)= − ￿ = − = − . z(z 1)(z +1) 2i(2i 1)(2i +1) 10 ￿z=2i − ￿ − By the Cauchy Residue Theorem, ￿ ￿ 3z3 +4z2 5z +1 i 3(1 + 2i) 7(2i 1) 34 15i − dz =2πi + + − + − =6πi. (z 2i)(z3 z) −2 10 10 10 ￿C − − ￿ ￿

31–4