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Flashcards: Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = \frac{e^{-z}}{z^3}$

in the region |z|=4 .

$3 \pi i$

$\pi i$

$- \frac{\pi i }{4}$

$- \frac{\pi i }{2}$

$- 2 \pi i$

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Flashcards: Residue Theory

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = \frac{e^{-z}}{z^3}$

in the region |z|=4 .

Your answer:

$3 \pi i$

Your answer:

$- \frac{\pi i }{4}$

Your answer:

$- \frac{\pi i }{2}$

Your answer:

$- 2 \pi i$

Correct answer:

$\pi i$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{z^3}$

a singularity exists where z= 0 . Thus, since where z=0 is the only singularity for f(z) inside |z|=3 , we seek to evaluate the residue for z=0 .

Observe,

$f(z) \newline = \frac{e^{-z}}{z^3} \newline = \frac{1}{z^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^{k-3}}{k!}$

The coefficient of $\frac{1}{z}$ is $\frac{(-1)^{2}}{2!} = \frac{1}{2}$ .

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{z^3}] = \frac{1}{2}$ .

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{z^2}dz = 2 \pi i \text{Res}_{z=0}[\frac{e^{-z}}{z^2}] = 2 \pi i (\frac{1}{2}) = \pi i$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{z^2} dz = \pi i$

Flashcards: Residue Theory

Cauchy's Residue Theorem is as follows:

Let be a simple closed contour, described positively. If a function is analytic inside except for a finite number of singular points $z_{k}$ inside , then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Using Cauchy's Residue Theorem, evaluate the integral of

$f(z) = \frac{e^{-z}}{(z-1)^3}$

in the region |z|=4

$2 \pi i$

$- 2 \pi i(1 + e)$

$-e^{2 \pi i}$

$\frac{ \pi i}{e}$

$- 2 \pi i$

Report an Error Copyright Notice

Flashcards: Residue Theory

Using Cauchy's Residue Theorem, evaluate the integral of

$f(z) = \frac{e^{-z}}{(z-1)^3}$

in the region |z|=4

Your answer:

$2 \pi i$

Your answer:

$- 2 \pi i(1 + e)$

Your answer:

$-e^{2 \pi i}$

Your answer:

$- 2 \pi i$

Correct answer:

$\frac{ \pi i}{e}$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{(z-1)^3}$

a singularity exists where z= 1 . Thus, since where z=1 is the only singularity for f(z) inside |z|=4 , we seek to evaluate the residue for z=1 .

Observe,

$f(z) \newline = \frac{e^{-z}}{(z-1)^3} \newline = \frac{e^{-u-1}}{u^3} \newline = \frac{ 1}{e} * \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^k}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^{k-3}}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{-1^{k}(z-1)^{k-3}}{k!}$

The coefficient of $\frac{1}{z-1}$ is $\frac{1}{e} * \frac{(-1)^{2}}{2!} = \frac{1}{2e}$ .

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{(z-1)^2}] = \frac{1}{2e}$ .

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{(z-1)^2}dz = 2 \pi i \text{Res}_{z=1}[\frac{e^{-z}}{(z-1)^2}] = 2 \pi i (\frac{1}{2e}) = \frac{ \pi i}{e}$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{(z-1)^2} dz = \frac{ \pi i}{e}$

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All Complex Analysis Resources

Flashcards: Residue Theory

Flashcards: Residue Theory

Flashcards: Residue Theory

Flashcards: Residue Theory

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Flashcards: Residue Theory

Flashcards: Residue Theory

Flashcards: Residue Theory

Flashcards: Residue Theory

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DMCA Complaint

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Complaint Details

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