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Topic 8 Notes
Jeremy Orloff
8 Residue Theorem
8.1 Poles and zeros
We remind you of the following terminology: Suppose f (z) is analytic at z 0 and
f (z) = an(z − z 0 )n^ + an+1(z − z 0 )n+1^ +... ,
with an 6 = 0. Then we say f has a zero of order n at z 0. If n = 1 we say z 0 is a simple zero.
Suppose f has an isolated singularity at z 0 and Laurent series
f (z) = bn (z − z 0 )n^
+... + b^1 z − z 0
- a 0 + a 1 (z − z 0 ) +...
which converges on 0 < |z − z 0 | < R and with bn 6 = 0. Then we say f has a pole of order n at z 0. If n = 1 we say z 0 is a simple pole.
There are several examples in the Topic 7 notes. Here is one more
Example 8.1.
f (z) = z^ + 1 z^3 (z^2 + 1) has isolated singularities at z = 0, ±i and a zero at z = −1. We will show that z = 0 is a pole of order 3, z = ±i are poles of order 1 and z = −1 is a zero of order 1. The style of argument is the same in each case.
At z = 0: f (z) =
z^3 ·^
z + 1 z^2 + 1. Call the second factor g(z). Since g(z) is analytic at z = 0 and g(0) = 1, it has a Taylor series g(z) =
z + 1 z^2 + 1 = 1 +^ a^1 z^ +^ a^2 z
Therefore f (z) =
z^3 +^
a 1 z^2 +^
a 2 z +^.... This shows z = 0 is a pole of order 3.
At z = i: f (z) = (^) z−^1 i · (^) z 3 z(+1z+i). Call the second factor g(z). Since g(z) is analytic at z = i, it has a Taylor series
g(z) = z^ + 1 z^3 (z + i)
= a 0 + a 1 (z − i) + a 2 (z − i)^2 +...
where a 0 = g(i) 6 = 0. Therefore
f (z) = a^0 z − i
This shows z = i is a pole of order 1.
The arguments for z = −i and z = −1 are similar.
1
8.2 Words: Holomorphic and meromorphic
Definition. A function that is analytic on a region A is called holomorphic on A.
A function that is analytic on A except for a set of poles of finite order is called meromorphic on A.
Example 8.2. Let
f (z) =
z + z^2 + z^3 (z − 2)(z − 3)(z − 4)(z − 5).
This is meromorphic on C with (simple) poles at z = 2, 3 , 4 , 5.
8.3 Behavior of functions near zeros and poles
The basic idea is that near a zero of order n, a function behaves like (z − z 0 )n^ and near a pole of order n, a function behaves like 1/(z − z 0 )n. The following make this a little more precise.
Behavior near a zero. If f has a zero of order n at z 0 then near z 0 ,
f (z) ≈ an(z − z 0 )n,
for some constant an.
Proof. By definition f has a Taylor series around z 0 of the form
f (z) = an(z − z 0 )n^ + an+1(z − z 0 )n+1^ +...
= an(z − z 0 )n
1 + an+ an
(z − z 0 ) + an+ an
(z − z 0 )^2 +...
Since the second factor equals 1 at z 0 , the claim follows.
Behavior near a finite pole. If f has a pole of order n at z 0 then near z 0 ,
f (z) ≈ bn (z − z 0 )n^
for some constant bn.
Proof. This is nearly identical to the previous argument. By definition f has a Laurent series around z 0 of the form
f (z) =
bn (z − z 0 )n^ +^
bn− 1 (z − z 0 )n−^1 +^...^ +^
b 1 z − z 0 +^ a^0 +^...
=
bn (z − z 0 )n
bn− 1 bn^ (z^ −^ z^0 ) +^
bn− 2 bn^ (z^ −^ z^0 )
Since the second factor equals 1 at z 0 , the claim follows.
8.3.1 Picard’s theorem and essential singularities
Near an essential singularity we have Picard’s theorem. We won’t prove or make use of this theorem in 18.04. Still, we feel it is pretty enough to warrant showing to you.
The residue of f at z 0 is b 1. This is denoted
Res(f, z 0 ) = b 1 or (^) zRes=z 0
f = b 1.
What is the importance of the residue? If γ is a small, simple closed curve that goes counterclockwise around b 1 then (^) ∫
γ
f (z) = 2πib 1.
γ small enough to be inside |z − z 0 | < r, surround z 0 and contain no other singularity of f.
This is easy to see by integrating the Laurent series term by term. The only nonzero integral comes from the term b 1 /z.
Example 8.5.
f (z) = e^1 /^2 z^ = 1 +^1 2 z
2(2z)^2
has an isolated singularity at 0. From the Laurent series we see that Res(f, 0) = 1/2.
Example 8.6.
(i) Let f (z) =
z^3 +
z^2 +
z + 5 + 6z. f has a pole of order 3 at z = 0 and Res(f, 0) = 4. (ii) Suppose f (z) =^2 z
where g is analytic at z = 0. Then, f has a simple pole at 0 and Res(f, 0) = 2. (iii) Let f (z) = cos(z) = 1 − z^2 /2! +.... Then f is analytic at z = 0 and Res(f, 0) = 0. (iv) Let f (z) = sin(z) z
=^1
z
z − z
3 3!
= 1 − z
2 3!
So, f has a removable singularity at z = 0 and Res(f, 0) = 0.
Example 8.7. Using partial fractions. Let
f (z) = (^) z 2 z+ 1.
Find the poles and residues of f.
Solution: Using partial fractions we write
f (z) =
z (z − i)(z + i) =
2 ·^
z − i +
2 ·^
z + i.
The poles are at z = ±i. We compute the residues at each pole:
At z = i: f (z) =^1 2
z − i
Therefore the pole is simple and Res(f, i) = 1/2.
At z = −i:
f (z) =
2 ·^
z + i + something analytic at^ −i.
Therefore the pole is simple and Res(f, −i) = 1/2.
Example 8.8. Mild warning! Let
f (z) = − 1 z(1 − z)
then we have the following Laurent expansions for f around z = 0.
On 0 < |z| < 1:
f (z) = −
z ·^
1 − z =^ −^
z (1 +^ z^ +^ z
Therefore the pole at z = 0 is simple and Res(f, 0) = −1.
On 1 < |z| < ∞:
f (z) = 1 z^2
1 − 1 /z
z^2
1 +^1
z
+^1
z^2
Even though this is a valid Laurent expansion you must not use it to compute the residue at 0. This is because the definition of residue requires that we use the Laurent series on the region 0 < |z − z 0 | < r.
Example 8.9. Let f (z) = log(1 + z).
This has a singularity at z = −1, but it is not isolated, so not a pole and therefore there is no residue at z = −1.
8.4.1 Residues at simple poles
Simple poles occur frequently enough that we’ll study computing their residues in some detail. Here are a number of ways to spot a simple pole and compute its residue. The justification for all of them goes back to Laurent series.
where a 1 = g′(z 0 ). So
1 g(z) =^
a 1 (z − z 0 )
1 + a a^21 (z − z 0 ) +...
The second factor on the right is analytic at z 0 and equals 1 at z 0. Therefore we know the Laurent expansion of 1/g is
1 g(z) =^
a 1 (z − z 0 ) (1 +^ c^1 (z^ −^ z^0 ) +^.. .)
Clearly the residue is 1/a 1 = 1/g′(z 0 ). QED.
Example 8.10. Let
f (z) = 2 +^ z^ +^ z
2 (z − 2)(z − 3)(z − 4)(z − 5)
Show all the poles are simple and compute their residues.
Solution: The poles are at z = 2, 3 , 4 , 5. They are all isolated. We’ll look at z = 2 the others are similar. Multiplying by z − 2 we get
g(z) = (z − 2)f (z) =
2 + z + z^2 (z − 3)(z − 4)(z − 5).
This is analytic at z = 2 and
g(2) = 8 − 6
So the pole is simple and Res(f, 2) = − 4 /3.
Example 8.11. Let
f (z) =
sin(z).
Find all the poles and their residues.
Solution: The poles of f (z) are the zeros of sin(z), i.e. nπ for n an integer. Since the derivative sin′(nπ) = cos(nπ) 6 = 0,
the zeros are simple and by Property 5 above
Res(f, nπ) = 1 cos(nπ)
= (−1)n.
Example 8.12. Let
f (z) = 1 z(z^2 + 1)(z − 2)^2
Identify all the poles and say which ones are simple.
Solution: Clearly the poles are at z = 0, ±i, 2.
At z = 0: g(z) = zf (z)
is analytic at 0 and g(0) = 1/4. So the pole is simple and the residue is g(0) = 1/4.
At z = i: g(z) = (z − i)f (z) = 1 z(z + i)(z − 2)^2
is analytic at i, the pole is simple and the residue is g(i).
At z = −i: This is similar to the case z = i. The pole is simple.
At z = 2: g(z) = (z − 2)f (z) = 1 z(z^2 + 1)(z − 2)
is not analytic at 2, so the pole is not simple. (It should be obvious that it’s a pole of order 2.)
Example 8.13. Let p(z), q(z) be analytic at z = z 0. Assume p(z 0 ) 6 = 0, q(z 0 ) = 0, q′(z 0 ) 6 = 0. Find
zRes=z 0
p(z) q(z)
Solution: Since q′(z 0 ) 6 = 0, q has a simple zero at z 0. So 1/q has a simple pole at z 0 and
Res(1/q, z 0 ) = (^) q′(^1 z 0 )
Since p(z 0 ) 6 = 0 we know
Res(p/q, z 0 ) = p(z 0 ) Res(1/q, z 0 ) = p(z^0 ) q′(z 0 )
8.4.2 Residues at finite poles
For higher-order poles we can make statements similar to those for simple poles, but the formulas and computations are more involved. The general principle is the following
Higher order poles. If f (z) has a pole of order k at z 0 then
g(z) = (z − z 0 )kf (z)
is analytic at z 0 and if g(z) = a 0 + a 1 (z − z 0 ) +...
then
Res(f, z 0 ) = ak− 1 =
g(k−1)(z 0 ) (k − 1)!.
Proof. This is clear using Taylor and Laurent series for g and f.
Example 8.14. Let
f (z) = sinh(z) z^5 and find the residue at z = 0.
Solution: We know the Taylor series for
sinh(z) = z + z^3 /3! + z^5 /5! +...
as claimed.
Sometimes we need more terms in the Laurent expansion of cot(z). There is no known easy formula for the terms, but we can easily compute as many as we need using the following technique.
Example 8.17. Compute the first several terms of the Laurent expansion of cot(z) around z = 0.
Solution: Since cot(z) has a simple pole at 0 we know
cot(z) =
b 1 z +^ a^0 +^ a^1 z^ +^ a^2 z
We also know
cot(z) = cos(z) sin(z)
=^1 −^ z
(^2) /2 + z (^4) /4! −... z − z^3 /3! + z^5 /5! −...
Cross multiplying the two expressions we get ( b 1 z +^ a^0 +^ a^1 z^ +^ a^2 z
z − z
3 3! +^
z^5 5! −^...
= 1 − z
2 2 +^
z^4 4! −^...
We can do the multiplication and equate the coefficients of like powers of z.
b 1 + a 0 z +
− b^1 3!
z^2 +
− a^0 3!
z^3 +
b 1 5!
− a^1 3!
z^4 = 1 − z
2 2!
4 4!
So, starting from b 1 = 1 and a 0 = 0, we get
−b 1 /3! + a 1 = − 1 /2! ⇒ a 1 = − 1 / 3 −a 0 /3! + a 2 = 0 ⇒ a 2 = 0 b 1 /5! − a 1 /3! + a 3 = 1/4! ⇒ a 3 = − 1 / 45.
As noted above, all the even terms are 0 as they should be. We have
cot(z) =^1 z
− z 3
− z
3 45
8.5 Cauchy Residue Theorem
This is one of the major theorems in 18.04. It will allow us to make systematic our previous somewhat ad hoc approach to computing integrals on contours that surround singularities.
Theorem. (Cauchy’s residue theorem) 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 ∫
C
f (z) dz = 2πi
residues of f inside C
Proof. The proof is based of the following figures. They only show a curve with two singu- larities inside it, but the generalization to any number of signularities 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’).
The left figure shows the curve C surrounding two poles z 1 and z 2 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˜ = C 1 + C 2 − C 3 − C 2 + C 4 + C 5 − C 6 − C 5
The left hand curve is C = C 1 + C 4. Since there are no poles inside C˜ we have, by Cauchy’s theorem, (^) ∫
C^ ˜
f (z) dz =
C 1 +C 2 −C 3 −C 2 +C 4 +C 5 −C 6 −C 5
f (z) dz = 0
Dropping C 2 and C 5 , which are both added and subtracted, this becomes ∫
C 1 +C 4
f (z) dz =
C 3 +C 6
f (z) dz (1)
If f (z) =... + b^2 (z − z 1 )^2
is the Laurent expansion of f around z 1 then ∫
C 3
f (z) dz =
C 3
... + b^2 (z − z 1 )^2
= 2πib 1 = 2πi Res(f, z 1 )
Likewise (^) ∫
C 6
f (z) dz = 2πi Res(f, z 2 ).
Using these residues and the fact that C = C 1 + C 4 , Equation 1 becomes ∫
C
f (z) dz = 2πi [Res(f, z 1 ) + Res(f, z 2 )].
That proves the residue theorem for the case of two poles. As we said, generalizing to any number of poles is straightforward.
Example 8.18. Let
f (z) = (^) z(z (^21) + 1).
Example 8.19. Compute (^) ∫
|z|=
5 z − 2 z(z − 1) dz.
Solution: Let f (z) = 5 z^ −^2 z(z − 1)
The poles of f are at z = 0, 1 and the contour encloses them both.
At z = 0: g(z) = zf (z) =
5 z − 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) =
5 z − 2 z is analytic at 1 so the pole is simple and
Res(f, 1) = g(1) = 3.
Finally (^) ∫
C
5 z − 2 z(z − 1) dz^ = 2πi^ [Res(f,^ 0) + Res(f,^ 1)] = 10πi.
Example 8.20. Compute (^) ∫
|z|=
z^2 sin(1/z) dz.
Solution: Let f (z) = z^2 sin(1/z).
f has an isolated singularity at z = 0. Using the Taylor series for sin(w) we get
z^2 sin(1/z) = z^2
z −^
3!z^3 +^
5!z^5 −^...
= z −
z +^...
So, Res(f, 0) = b 1 = − 1 /6. Thus the residue theorem gives ∫
|z|=
z^2 sin(1/z) dz = 2πi Res(f, 0) = − iπ 3
Example 8.21. Compute (^) ∫
C
dz z(z − 2)^4 dz,
where, C : |z − 2 | = 1.
Solution: Let f (z) = 1 z(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:
1 z
2 + (z − 2)
2 ·^
1 + (z − 2)/ 2
=
z − 2 2 +
(z − 2)^2 4 −^
(z − 2)^3 8 +^...
This is valid on 0 < |z − 2 | < 2. So,
f (z) = 1 (z − 2)^4
z
2(z − 2)^4
4(z − 2)^3
8(z − 2)^2
16(z − 2)
Thus, Res(f, 2) = − 1 /16 and ∫
C
f (z) dz = 2πi Res(f, 2) = −
πi
Example 8.22. Compute (^) ∫
C
sin(z) dz
over the contour C shown.
We should first explain the idea here. The interior of a simple closed curve is everything to left as you traverse the curve. The curve C is oriented counterclockwise, so its interior contains all the poles of f. The residue theorem says the integral over C is determined by the residues of these poles.
On the other hand, the interior of the curve −C is everything outside of C. There are no poles of f in that region. If we want the residue theorem to hold (which we do –it’s that important) then the only option is to have a residue at ∞ and define it as we did.
The definition of the residue at infinity assumes all the poles of f are inside C. Therefore the residue theorem implies
Res(f, ∞) = −
the residues of f.
To make this useful we need a way to compute the residue directly. This is given by the following theorem.
Theorem. If f is analytic in C except for a finite number of singularities then
Res(f, ∞) = − Res
w^2
f (1/w), 0
Proof. The proof is just a change of variables: w = 1/z.
Change of variable: w = 1/z
First note that z = 1/w and dz = −(1/w^2 ) dw.
Next, note that the map w = 1/z carries the positively oriented z-circle of radius R to the negatively oriented w-circle of radius 1/R. (To see the orientiation, follow the circled points 1, 2, 3, 4 on C in the z-plane as they are mapped to points on C˜ in the w-plane.) Thus,
Res(f, ∞) = − 1 2 πi
C
f (z) dz = 1 2 πi
C^ ˜
f (1/w)^1 w^2
dw
Finally, note that z = 1/w maps all the poles inside the circle C to points outside the circle C˜. So the only possible pole of (1/w^2 )f (1/w) that is inside C˜ is at w = 0. Now, since C˜ is oriented clockwise, the residue theorem says
1 2 πi
C^ ˜
f (1/w)^1 w^2
dw = − Res
w^2
f (1/w), 0
Comparing this with the equation just above finishes the proof.
Example 8.23. Let
f (z) = (^) z^5 (zz^ −−^2 1).
Earlier we computed (^) ∫
|z|=
f (z) dz = 10πi
by computing residues at z = 0 and z = 1. Recompute this integral by computing a single residue at infinity.
Solution: 1 w^2
f (1/w) = 1 w^2
5 /w − 2 (1/w)(1/w − 1)
= 5 −^2 w w(1 − w)
We easily compute that
Res(f, ∞) = − Res
w^2
f (1/w), 0
Since |z| = 2 contains all the singularities of f we have ∫
|z|=
f (z) dz = − 2 πi Res(f, ∞) = 10πi.
This is the same answer we got before!
