5.3 Residue Theorem

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We use the Residue Theorem to compute integrals of complex functions around closed contours.

Recall that the coefficient $c_{-1}$ in the Laurent series of a function centered at $z_0$ , $f(z) = \sum _{k=-\infty }^\infty c_k (z-z_0)^k$ is called the residue of $f$ at $z_0$ , denoted Res $(f,z_0)$ .

If we integrated this Laurent series term by term around a positively oriented simple closed curve, $C$ , containing no other singularities in its interior besides (possibly) $z_0$ then we would obtain: \begin{align*} \int _C \left (\sum _{k=-\infty }^\infty c_k (z-z_0)^k \right )\, dz &= \sum _{k=-\infty }^\infty \left ( \int _C c_k (z-z_0)^k \, dz \right )\\ &= c_{-1} \int _C \frac{1}{z-z_0} \, dz \quad \text{(verify)}\\ &= 2\pi i c_{-1} \quad \text{(verify)}\\ &= 2\pi i \text{Res}(f, z_0) \end{align*}

Combining the above result with the Extended Deformation of Contour Theorem, we obtain the Residue Theorem:

Residue Theorem Let $C$ be a positively oriented, simple closed contour and suppose the function $f$ is analytic along $C$ and inside $C$ , except at finitely many isolated singularities at $z_1, z_2, \dots , z_n$ . Then $\int _C f(z) \, dz = 2\pi i \sum _{k=1}^n \text {Res}(f,z_k)$

example 1a Let $C$ be a positively oriented, simple closed contour containing the origin in its interior. Compute the contour integral $\int _C e^{1/z} \, dz$ The integrand has a singularity at the origin (and nowhere else) and its Laurent series centered at $0$ is $e^{1/z} = 1 + \frac {1}{z} + \frac {1}{2!z^2} + \frac {1}{3! z^3} + \cdots$ Hence, Res $(e^{1/z},0) = c_{-1} = 1$ . By the Residue Theorem we obtain $\int _C e^{1/z} \, dz = 2\pi i \text {Res}(e^{1/z}, 0) = 2\pi i$

example 1b Let $C$ be a positively oriented, simple closed contour containing the points $0$ and $1$ in its interior. Compute the contour integral $\int _C \frac {e^{1/z}}{1-z} \, dz$

The function $\dis f(z) = \frac {e^{1/z}}{1-z}$ has singularities at both $0$ and $1$ , we need to find Laurent series for $f$ centered at both of these points. At $z = 1$ , the function $e^{1/z}$ is analytic an hence has a Taylor series centered at $1$ of the form: $e^{1/z} = c_0 + c_1(z-1) + c_2(z-1)^2 + \cdots$ The Laurent series for $\frac {1}{1-z}$ centered at $1$ is simply $\frac {1}{1-z} = -\frac {1}{z-1}$ Multiplying these Taylor and Laurent series gives the Laurent series for $f$ : $\frac {e^{1/z}}{1-z} = -\frac {c_0}{z-1} -c_1 -c_2(z-1) - c_3(z-1)^2 - \cdots$ Thus Res $(f, 1) = -c_0 = -e$ since the is the constant term of the Taylor series of $e^{1/z} =$ centered at $1$ is $e^{1/1} =e$ .
Next, at $z = 0$ , the function $\frac {1}{1-z}$ is analytic and has Taylor series $\frac {1}{1-z} = 1 + z + z^2 + z^3 + \cdots$ and the function $e^{1/z}$ has Laurent series $e^{1/z} = 1 + \frac {1}{z} + \frac {1}{2! z^2} + \frac {1}{3! z^3} + \cdots$ The Laurent series of $f$ is obtained by multiplying these Taylor and Laurent series together $\frac {e^{1/z}}{1-z} = \left (1 + z + z^2 + z^3 + \cdots \right ) \cdot \left ( 1 + \frac {1}{z} + \frac {1}{2! z^2} + \frac {1}{3! z^3} + \cdots \right )$ We only need the residue of $f$ , so let’s compute the coefficient of $\frac {1}{z}$ in the product above. We get: $\text {Res}(f, 0) = 1 + \frac {1}{2!} +\frac {1}{3!} + \frac {1}{4!} + \cdots = e-1$ Finally, we can evaluate the contour integral using the Residue Theorem: $\int _C \frac {e^{1/z}}{1-z} \, dz = 2\pi i \left [\text {Res}(f,0) + \text {Res}(f,1)\right ] = -2\pi i$

(problem 1a) Let $C$ be a positively oriented, simple closed curve containing the origin in its interior. Compute $\int _C \sin \left (\frac {1}{z}\right ) \, dz = \answer {2\pi i}$

(problem 1b) Let $C$ be a positively oriented, simple closed curve containing the origin in its interior. Compute $\int _C z^2\sin \left (\frac {1}{z}\right ) \, dz = \answer {-\pi i/3}$

(problem 1c) Let $C$ be a positively oriented, simple closed curve containing the origin in its interior. Compute $\int _C z^3\cos \left (\frac {1}{z}\right ) \, dz = \answer {\pi i/(12)}$

(problem 1d) Let $C$ be a positively oriented, simple closed curve containing $0$ and $1$ in its interior. Compute $\int _C f(z) \, dz \quad \text {where} \quad f(z) = \frac {\sin \left (\frac {1}{z}\right )}{1-z}$

Res $(f, 0) = \answer {\sin (1)}$ Res $(f, 1) = \answer {-\sin (1)}$
$\int _C \frac {\sin \left (\frac {1}{z}\right )}{1-z} \, dz = \answer {0}$

Here is a video solution to problem 1c:

Here is a video solution to problem 1d:

2024-09-27 14:07:17