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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.
Theorem \(\PageIndex{1}\) 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
\[\int_{C} f(z) \ dz = 2\pi i \sum \text{ residues of } f \text{ inside } C \nonumber \]
- 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’).
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
\[\tilde{C} = C_1 + C_2 - C_3 - C_2 + C_4 + C_5 - C_6 - C_5 \nonumber \]
The left hand curve is \(C = C_1 + C_4\). Since there are no poles inside \(\tilde{C}\) we have, by Cauchy’s theorem,
\[\int_{\tilde{C}} f(z) \ dz = \int_{C_1 + C_2 - C_3 - C_2 + C_4 + C_5 - C_6 - C_5} f(z) \ dz = 0 \nonumber \]
Dropping \(C_2\) and \(C_5\), which are both added and subtracted, this becomes
\[\int_{C_1 + C_4} f(z)\ dz = \int_{C_3 + C_6} f(z)\ dz \nonumber \]
If
\[f(z) = \ ... + \dfrac{b_2}{(z - z_1)^2} + \dfrac{b_1}{z - z_1} + a_0 + a_1 (z - z_1) + \ ... \nonumber \]
is the Laurent expansion of \(f\) around \(z_1\) then
\[\begin{array} {rcl} {\int_{C_3} f(z)\ dz} & = & {\int_{C_3}\ ... + \dfrac{b_2}{(z - z_1)^2} + \dfrac{b_1}{z - z_1} + a_0 + a_1 (z - z_1) + \ ... dz} \\ {} & = & {2\pi i b_1} \\ {} & = & {2\pi i \text{Res} (f, z_1)} \end{array} \nonumber \]
Likewise
\[\int_{C_6} f(z)\ dz = 2\pi i \text{Res} (f, z_2). \nonumber \]
Using these residues and the fact that \(C = C_1 + C_4\), Equation 9.5.4 becomes
\[\int_C f(z)\ dz = 2\pi i [\text{Res} (f, z_1) + \text{Res} (f, z_2)]. \nonumber \]
That proves the residue theorem for the case of two poles. As we said, generalizing to any number of poles is straightforward.
Example \(\PageIndex{1}\)
Let
\[f(z) = \dfrac{1}{z(z^2 + 1)}. \nonumber \]
Compute \(\int f(z)\ dz\) over each of the contours \(C_1, C_2, C_3, C_4\) shown.
Solution
The poles of \(f(z)\) are at \(z = 0, \pm i\). Using the residue theorem we just need to compute the residues of each of these poles.
At \(z = 0\):
\[g(z) = zf(z) = \dfrac{1}{z^2 + 1} \nonumber \]
is analytic at 0 so the pole is simple and
\[\text{Res} (f, 0) = g(0) = 1. \nonumber \]
At \(z = i\):
\[g(z) = (z - i) f(z) = \dfrac{1}{z(z + i)} \nonumber \]
is analytic at \(i\) so the pole is simple and
\[\text{Res} (f, i) = g(i) = -1/2. \nonumber \]
At \(z = -i\):
\[g(z) = (z + i) f(z) = \dfrac{1}{z (z - i)} \nonumber \]
is analytic at \(-i\) so the pole is simple and
\[\text{Res} (f, -i) = g(-i) = -1/2. \nonumber \]
Using the residue theorem we have
\[\begin{array} {l} {\int_{C_1} f(z)\ dz = 0 \text{ (since } f \text{ is analytic inside } C_1)} \\ {\int_{C_2} f(z)\ dz = 2 \pi i \text{Res} (f, i) = -\pi i} \\ {\int_{C_3} f(z)\ dz = 2\pi i [\text{Res}(f, i) + \text{Res} (f, 0)] = \pi i} \\ {\int_{C_4} f(z)\ dz = 2\pi i [\text{Res} (f, i) + \text{Res} (f, 0) + \text{Res} (f, -i)] = 0.} \end{array} \nonumber \]
Example \(\PageIndex{2}\)
Compute
\[\int_{|z| = 2} \dfrac{5z - 2}{z (z - 1)}\ dz. \nonumber \]
Solution
Let
\[f(z) = \dfrac{5z - 2}{z(z - 1)}. \nonumber \]
The poles of \(f\) are at \(z = 0, 1\) and the contour encloses them both.
At \(z = 0\):
\[g(z) = zf(z) = \dfrac{5z - 2}{(z - 1)} \nonumber \]
is analytic at 0 so the pole is simple and
\[\text{Res} (f, 0) = g(0) = 2. \nonumber \]
At \(z = 1\):
\[g(z) = (z - 1) f(z) = \dfrac{5z - 2}{z} \nonumber \]
is analytic at 1 so the pole is simple and
\[\text{Res} (f, 1) = g(1) = 3. \nonumber \]
Finally
\[\int_{C} \dfrac{5z - 2}{z(z - 1)} \ dz = 2\pi i [\text{Res} (f, 0) + \text{Res} (f, 1)] = 10 \pi i. \nonumber \]
Example \(\PageIndex{3}\)
Compute
\[\int_{|z| = 1} z^2 \sin (1/z)\ dz. \nonumber \]
Solution
Let
\[f(z) = z^2 \sin (1/z). \nonumber \]
\(f\) has an isolated singularity at \(z = 0\). Using the Taylor series for \(\sin (w)\) we get
\[z^2 \sin (1/z) = z^2 \left(\dfrac{1}{z} - \dfrac{1}{3! z^3} + \dfrac{1}{5! z^5} - \ ... \right) = z - \dfrac{1/6}{z} + \ ... \nonumber \]
So, \(\text{Res} (f, 0) = b_1 = -1/6\). Thus the residue theorem gives
\[\int_{|z| = 1} z^2 \sin (1/z)\ dz = 2\pi i \text{Res} (f, 0) = - \dfrac{i \pi}{3}. \nonumber \]
Example \(\PageIndex{4}\)
Compute
\[\int_C \dfrac{dz}{z(z - 2)^4} \ dz, \nonumber \]
where, \(C : |z - 2| = 1\).
Solution
Let
\[f(z) = \dfrac{1}{z(z - 2)^4}. \nonumber \]
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:
\[\begin{array} {rcl} {\dfrac{1}{z}} & = & {\dfrac{1}{2 + (z - 2)}} \\ {} & = & {\dfrac{1}{2} \cdot \dfrac{1}{1 + (z - 2)/2}} \\ {} & = & {\dfrac{1}{2} (1 - \dfrac{z - 2}{2} + \dfrac{(z - 2)^2}{4} - \dfrac{(z - 2)^3}{8} + \ ..)} \end{array} \nonumber \]
This is valid on \(0 < |z - 2| < 2\). So,
\[f(z) = \dfrac{1}{(z - 4)^4} \cdot \dfrac{1}{z} = \dfrac{1}{2(z - 2)^4} - \dfrac{1}{4(z - 2)^3} + \dfrac{1}{8(z - 2)^2} - \dfrac{1}{16(z - 2)} + \ ... \nonumber \]
Thus, \(\text{Res} (f, 2) = -1/16\) and
\[\int_C f(z)\ dz = 2\pi i \text{Res} (f, 2) = - \dfrac{\pi i}{8}. \nonumber \]
Example \(\PageIndex{5}\)
Compute
\[\int_C \dfrac{1}{\sin (z)} \ dz \nonumber \]
over the contour \(C\) shown.
Solution
Let
\[f(z) = 1/ \sin (z). \nonumber \]
There are 3 poles of \(f\) inside \(C\) at \(0, \pi\) and \(2\pi\). We can find the residues by taking the limit of \((z - z_0) f(z)\). 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 \(z = 0\):
\[\lim_{z \to 0} \dfrac{z}{\sin (z)} = \lim_{z \to 0} \dfrac{1}{\cos (z)} = 1. \nonumber \]
Since the limit exists, \(z = 0\) is a simple pole and
\[\text{Res} (f, 0) = 1. \nonumber \]
At \(z = \pi\):
\[\lim_{z \to \pi} \dfrac{z - \pi}{\sin (z)} = \lim_{z \to \pi} \dfrac{1}{\cos (z)} = -1. \nonumber \]
Since the limit exists, \(z = \pi\) is a simple pole and
\[\text{Res} (f, \pi) = -1. \nonumber \]
At \(z = 2 \pi\): The same argument shows
\[\text{Res} (f, 2\pi) = 1. \nonumber \]
Now, by the residue theorem
\[\int_C f(z)\ dz = 2\pi i [\text{Res} (f, 0) + \text{Res} (f, \pi) + \text{Res} (f, 2\pi)] = 2\pi i. \nonumber \]
