9.5: Cauchy Residue Theorem
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This is one of the major theorems in complex analysis and will allow us to make systematic our previous somewhat ad hoc approach to computing integrals on contours that surround singularities.
Suppose f(z) is analytic in the region A except for a set of isolated singularities. Also suppose C is a simple closed curve in A that doesn’t go through any of the singularities of f and is oriented counterclockwise. Then
∫Cf(z) dz=2πi∑ residues of f inside C
- Proof
-
The proof is based of the following figures. They only show a curve with two singularities inside it, but the generalization to any number of singularities is straightforward. In what follows we are going to abuse language and say pole when we mean isolated singularity, i.e. a finite order pole or an essential singularity (‘infinite order pole’).
Figure 9.5.1: Applying the Cauchy residue theorem. (CC BY-NC; Ümit Kaya) The left figure shows the curve C surrounding two poles z1 and z2 of f. The right figure shows the same curve with some cuts and small circles added. It is chosen so that there are no poles of f inside it and so that the little circles around each of the poles are so small that there are no other poles inside them. The right hand curve is
˜C=C1+C2−C3−C2+C4+C5−C6−C5
The left hand curve is C=C1+C4. Since there are no poles inside ˜C we have, by Cauchy’s theorem,
∫˜Cf(z) dz=∫C1+C2−C3−C2+C4+C5−C6−C5f(z) dz=0
Dropping C2 and C5, which are both added and subtracted, this becomes
∫C1+C4f(z) dz=∫C3+C6f(z) dz
If
f(z)= ...+b2(z−z1)2+b1z−z1+a0+a1(z−z1)+ ...
is the Laurent expansion of f around z1 then
∫C3f(z) dz=∫C3 ...+b2(z−z1)2+b1z−z1+a0+a1(z−z1)+ ...dz=2πib1=2πiRes(f,z1)
Likewise
∫C6f(z) dz=2πiRes(f,z2).
Using these residues and the fact that C=C1+C4, Equation 9.5.4 becomes
∫Cf(z) dz=2πi[Res(f,z1)+Res(f,z2)].
That proves the residue theorem for the case of two poles. As we said, generalizing to any number of poles is straightforward.
Let
f(z)=1z(z2+1).
Compute ∫f(z) dz over each of the contours C1,C2,C3,C4 shown.
Solution
The poles of f(z) are at z=0,±i. Using the residue theorem we just need to compute the residues of each of these poles.
At z=0:
g(z)=zf(z)=1z2+1
is analytic at 0 so the pole is simple and
Res(f,0)=g(0)=1.
At z=i:
g(z)=(z−i)f(z)=1z(z+i)
is analytic at i so the pole is simple and
Res(f,i)=g(i)=−1/2.
At z=−i:
g(z)=(z+i)f(z)=1z(z−i)
is analytic at −i so the pole is simple and
Res(f,−i)=g(−i)=−1/2.
Using the residue theorem we have
∫C1f(z) dz=0 (since f is analytic inside C1)∫C2f(z) dz=2πiRes(f,i)=−πi∫C3f(z) dz=2πi[Res(f,i)+Res(f,0)]=πi∫C4f(z) dz=2πi[Res(f,i)+Res(f,0)+Res(f,−i)]=0.
Compute
∫|z|=25z−2z(z−1) dz.
Solution
Let
f(z)=5z−2z(z−1).
The poles of f are at z=0,1 and the contour encloses them both.
At z=0:
g(z)=zf(z)=5z−2(z−1)
is analytic at 0 so the pole is simple and
Res(f,0)=g(0)=2.
At z=1:
g(z)=(z−1)f(z)=5z−2z
is analytic at 1 so the pole is simple and
Res(f,1)=g(1)=3.
Finally
∫C5z−2z(z−1) dz=2πi[Res(f,0)+Res(f,1)]=10πi.
Compute
∫|z|=1z2sin(1/z) dz.
Solution
Let
f(z)=z2sin(1/z).
f has an isolated singularity at z=0. Using the Taylor series for sin(w) we get
z2sin(1/z)=z2(1z−13!z3+15!z5− ...)=z−1/6z+ ...
So, Res(f,0)=b1=−1/6. Thus the residue theorem gives
∫|z|=1z2sin(1/z) dz=2πiRes(f,0)=−iπ3.
Compute
∫Cdzz(z−2)4 dz,
where, C:|z−2|=1.
Solution
Let
f(z)=1z(z−2)4.
The singularity at z=0 is outside the contour of integration so it doesn’t contribute to the integral. To use the residue theorem we need to find the residue of f at z=2. There are a number of ways to do this. Here’s one:
1z=12+(z−2)=12⋅11+(z−2)/2=12(1−z−22+(z−2)24−(z−2)38+ ..)
This is valid on 0<|z−2|<2. So,
f(z)=1(z−4)4⋅1z=12(z−2)4−14(z−2)3+18(z−2)2−116(z−2)+ ...
Thus, Res(f,2)=−1/16 and
∫Cf(z) dz=2πiRes(f,2)=−πi8.
Compute
∫C1sin(z) dz
over the contour C shown.
Solution
Let
f(z)=1/sin(z).
There are 3 poles of f inside C at 0,π and 2π. We can find the residues by taking the limit of (z−z0)f(z). Each of the limits is computed using L’Hospital’s rule. (This is valid, since the rule is just a statement about power series. We could also have used Property 5 from the section on residues of simple poles above.)
At z=0:
limz→0zsin(z)=limz→01cos(z)=1.
Since the limit exists, z=0 is a simple pole and
Res(f,0)=1.
At z=π:
limz→πz−πsin(z)=limz→π1cos(z)=−1.
Since the limit exists, z=π is a simple pole and
Res(f,π)=−1.
At z=2π: The same argument shows
Res(f,2π)=1.
Now, by the residue theorem
∫Cf(z) dz=2πi[Res(f,0)+Res(f,π)+Res(f,2π)]=2πi.