Residue Theory - Complex Analysis

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Question

Find the residue at for the function

$f(z) = $$\frac{4}{z+z^2$}$$ .

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Answer

Observe,

$f(z) \newline = $\frac{4}{z + $z^2$}$ \newline =\frac{4}{z}$ * $\frac{1}{z+1}$ \newline = $\frac{4}{z}$ \displaystyle\sum_${k=0}^{\infty}$ $(-1)^{k}$ $z^{k}$ \newline = 4\displaystyle\sum_${k=0}^{\infty}$ $(-1)^{k}$ $z^{k -1}$ \newline = 4($\frac{1}{z}$ -1 + z - $z^2$ $+z^3$ + \cdots)$

The coefficient of $\frac{1}{z}$ is .

Thus,

.

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Question

Find the residue at of

$f(z) = \cos ($\frac{1}{z}$)$ .

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Answer

Let $u = $\frac{1}{z}$$ .

Observe,

$f(z) \newline = \cos ($\frac{1}{z}$) \newline =\cos (u) \newline = \displaystyle\sum_${k=0}^{\infty}$ $$\frac{(-1)^{n}$$}{(2n)!} $u^{2n}$ \newline = \displaystyle\sum_${k=0}^{\infty}$ $$\frac{(-1)^{n}$$}{(2n)!} $z^{-2n}$$

The coefficient of $\frac{1}{z}$ is since there is no $\frac{1}{z}$ term in the sum.

Thus,

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Question

Find the residue at of the function

$f(z) = $$\frac{z^2$$-sin(z)}{z^2$}$$ .

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Answer

Observe,

$f(z) \newline = $\frac{z-sin(z)}{z}$ \newline = 1 - $$\frac{sin(z)}{z^2$}$ \newline = 1 - \displaystyle\sum_${k=0}^{\infty}$ $$\frac{-1^{k}$$}{(2n+1)!} $z^{2n-1}$$ .

The coefficient of $\frac{1}{z}$ is .

Thus,

.

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Question

Find the residue of the function

$f(z)= $$\frac{cot(z)}{z^6$}$$ .

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Answer

Observe

$f(z) \newline = $$\frac{cot(z)}{z^6$}$ \newline = $$\frac{1}{z^6$}$ ($\frac{1}{z}$ - $\frac{1}{3}$ z - $\frac{1}{45}$ $z^3$ - $\frac{2}{945}$ $z^5$ - \cdots) \newline = $$\frac{1}{z^7$}$ - $\frac{1}{3}$ $$\frac{1}{z^5$}$ - $\frac{1}{45}$ $$\frac{1}{z^3$}$ - $\frac{2}{945}$ $\frac{1}{z}$ - \cdots$

The coefficient of $\frac{1}{z}$ is $- $\frac{2}{945}$$ .

Thus,

.

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Question

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 inside , then

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 .

Tap to reveal answer

Answer

Note, for

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

a singularity exists where . Thus, since where is the only singularity for inside , we seek to evaluate the residue for .

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 .

Thus,

.

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,

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Question

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 inside , then

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

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Answer

Note, for

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

a singularity exists where . Thus, since where is the only singularity for inside , we seek to evaluate the residue for .

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,

.

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,

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Question

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 inside , then

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

in the region .

Tap to reveal answer

Answer

Note, there is one singularity for where .

Let

$u = $\frac{1}{z}$$

Then

$z = $\frac{1}{u}$$

so

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

Therefore, there is one singularity for where . Hence, we seek to compute the residue for where

Observe,

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

So, when , .

Thus, the coefficient of $\frac{1}{z}$ is $\frac{1}{4!}$ .

Therefore,

Hence, by Cauchy's Residue Theorem,

$\displaystyle $\int_{|z|=3}$ $z^{2}$ $e^{$\frac{1}${z}$} dz = 2 \pi i $\text{Res}$_${z=0}[z^2$ $e^{$\frac{1}${z}$}] = 2 \pi i * $\frac{1}{4!}$ = $\frac{\pi i}{12}$$

Therefore,

$\displaystyle $\int_{|z|=3}$ $z^{2}$ $e^{$\frac{1}${z}$} dz = $\frac{\pi i}{12}$$

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

$f(z) = 5$\frac{z+1}{z(z-2)}$$

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe,

The coefficient of $\frac{1}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

$\displaystyle $\int_{|z|=3}$ f(z) dz = 2 \pi i $\text{Res}$_${z=0}[$\frac{1}{z^2$}$f($\frac{1}{z}$)] = 2 \pi i (5) = 10 \pi i$

Hence,

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

$f(z) = $$\frac{z^6$$}{1-z^2$}$$

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe,

The coefficient of $\frac{1}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

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

Hence,

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

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

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe,

The coefficient of $\frac{1}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

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

Hence,

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

$f(z) = $\frac{2}{z}$$

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe, the coefficient of $\frac{2}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

$\displaystyle $\int_{|z|=2}$ f(z) dz = 2 \pi i $\text{Res}$_${z=0}[$\frac{1}{z^2$}$f($\frac{1}{z}$)] = 2 \pi i (2) = 4 \pi i$

Hence,

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

$f(z) = $\frac{7z-3}{z(z-1)}$$

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe,

The coefficient of $\frac{1}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

$\displaystyle $\int_{|z|=2}$ f(z) dz = 2 \pi i $\text{Res}$_${z=0}[$\frac{1}{z^2$}$f($\frac{1}{z}$)] = 2 \pi i (7) = 14 \pi i$

Hence,

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Question

Find the residue at for the function

$f(z) = $$\frac{4}{z+z^2$}$$ .

Tap to reveal answer

Answer

Observe,

$f(z) \newline = $\frac{4}{z + $z^2$}$ \newline =\frac{4}{z}$ * $\frac{1}{z+1}$ \newline = $\frac{4}{z}$ \displaystyle\sum_${k=0}^{\infty}$ $(-1)^{k}$ $z^{k}$ \newline = 4\displaystyle\sum_${k=0}^{\infty}$ $(-1)^{k}$ $z^{k -1}$ \newline = 4($\frac{1}{z}$ -1 + z - $z^2$ $+z^3$ + \cdots)$

The coefficient of $\frac{1}{z}$ is .

Thus,

.

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Question

Find the residue at of

$f(z) = \cos ($\frac{1}{z}$)$ .

Tap to reveal answer

Answer

Let $u = $\frac{1}{z}$$ .

Observe,

$f(z) \newline = \cos ($\frac{1}{z}$) \newline =\cos (u) \newline = \displaystyle\sum_${k=0}^{\infty}$ $$\frac{(-1)^{n}$$}{(2n)!} $u^{2n}$ \newline = \displaystyle\sum_${k=0}^{\infty}$ $$\frac{(-1)^{n}$$}{(2n)!} $z^{-2n}$$

The coefficient of $\frac{1}{z}$ is since there is no $\frac{1}{z}$ term in the sum.

Thus,

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Question

Find the residue at of the function

$f(z) = $$\frac{z^2$$-sin(z)}{z^2$}$$ .

Tap to reveal answer

Answer

Observe,

$f(z) \newline = $\frac{z-sin(z)}{z}$ \newline = 1 - $$\frac{sin(z)}{z^2$}$ \newline = 1 - \displaystyle\sum_${k=0}^{\infty}$ $$\frac{-1^{k}$$}{(2n+1)!} $z^{2n-1}$$ .

The coefficient of $\frac{1}{z}$ is .

Thus,

.

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Question

Find the residue of the function

$f(z)= $$\frac{cot(z)}{z^6$}$$ .

Tap to reveal answer

Answer

Observe

$f(z) \newline = $$\frac{cot(z)}{z^6$}$ \newline = $$\frac{1}{z^6$}$ ($\frac{1}{z}$ - $\frac{1}{3}$ z - $\frac{1}{45}$ $z^3$ - $\frac{2}{945}$ $z^5$ - \cdots) \newline = $$\frac{1}{z^7$}$ - $\frac{1}{3}$ $$\frac{1}{z^5$}$ - $\frac{1}{45}$ $$\frac{1}{z^3$}$ - $\frac{2}{945}$ $\frac{1}{z}$ - \cdots$

The coefficient of $\frac{1}{z}$ is $- $\frac{2}{945}$$ .

Thus,

.

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Question

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 inside , then

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 .

Tap to reveal answer

Answer

Note, for

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

a singularity exists where . Thus, since where is the only singularity for inside , we seek to evaluate the residue for .

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 .

Thus,

.

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,

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Question

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 inside , then

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

Tap to reveal answer

Answer

Note, for

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

a singularity exists where . Thus, since where is the only singularity for inside , we seek to evaluate the residue for .

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,

.

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,

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Question

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 inside , then

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

in the region .

Tap to reveal answer

Answer

Note, there is one singularity for where .

Let

$u = $\frac{1}{z}$$

Then

$z = $\frac{1}{u}$$

so

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

Therefore, there is one singularity for where . Hence, we seek to compute the residue for where

Observe,

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

So, when , .

Thus, the coefficient of $\frac{1}{z}$ is $\frac{1}{4!}$ .

Therefore,

Hence, by Cauchy's Residue Theorem,

$\displaystyle $\int_{|z|=3}$ $z^{2}$ $e^{$\frac{1}${z}$} dz = 2 \pi i $\text{Res}$_${z=0}[z^2$ $e^{$\frac{1}${z}$}] = 2 \pi i * $\frac{1}{4!}$ = $\frac{\pi i}{12}$$

Therefore,

$\displaystyle $\int_{|z|=3}$ $z^{2}$ $e^{$\frac{1}${z}$} dz = $\frac{\pi i}{12}$$

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Question

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 inside , then

For the following problem, use a modified version of the theorem which goes as follows:

Residue Theorem

If a function is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour , then

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

Use the Residue Theorem to evaluate the integral of

$f(z) = 5$\frac{z+1}{z(z-2)}$$

in the region .

Tap to reveal answer

Answer

Note,

Thus, seeking to apply the Residue Theorem above for inside , we evaluate the residue for .

Observe,

The coefficient of $\frac{1}{z}$ is .

Thus,

.

Therefore, by the Residue Theorem above,

$\displaystyle $\int_{|z|=3}$ f(z) dz = 2 \pi i $\text{Res}$_${z=0}[$\frac{1}{z^2$}$f($\frac{1}{z}$)] = 2 \pi i (5) = 10 \pi i$

Hence,

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