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Want to review Complex Analysis but don’t feel like sitting for a whole test at the moment? Varsity Tutors has you covered with thousands of different Complex Analysis flashcards! Our Complex Analysis flashcards allow you to practice with as few or as many questions as you like. Get some studying in now with our numerous Complex Analysis flashcards.

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 .

$\pi i$

$3 \pi i$

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

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

$- 2 \pi i$

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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:

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

Your answer:

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

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

$- 2 \pi i$

$2 \pi i$

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

$-e^{2 \pi i}$

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

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:

$2 \pi i$

Your answer:

$-e^{2 \pi i}$

Your answer:

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

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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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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