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Complex Analysis : Complex Analysis

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Example Questions

Example Question #4 : 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

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

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

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

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 |z|=3 .

Possible Answers:

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

$\pi i$

$- \frac{\pi i}{3}$

$2 \pi i$

Correct answer:

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{\frac{1}{z^6}}{1 - \frac{1}{z^2}} = \frac{\frac{1}{z^6}}{z^2 - 1} = \frac{1}{z^8 - z^ 6} = -\frac{1}{z^6}* \frac{1}{1-z^2}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = -\frac{1}{z^6}* \frac{1}{1-z^2} \newline = -\frac{1}{z^6} \displaystyle \sum_{k=0}^{\infty} z^{2k} \newline = - \displaystyle \sum_{k=0}^{\infty} z^{2k -6}$

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

Thus,

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

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,

$\int_{|z|=3} f(z)dz =0$

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Example Question #1 : 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

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

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

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

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 |z|=4 .

Possible Answers:

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

$- 2 \pi i$

$10 \pi i$

$\pi i$

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

Correct answer:

$10 \pi i$

Explanation:

Note,

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

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=4 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{5}{z} (\frac{1}{1 - 2z}) + \frac{5}{1 - 2z} \newline = \frac{5}{z} \displaystyle \sum_{k=0}^{\infty} 2^{k} z^k + 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^{k-1} + 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle \sum_{k=0}^{\infty} 2^{k}( z^{k-1} + z^k)$

The coefficient of $\frac{1}{z}$ is 5 *2^0 = 5 .

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 5$ .

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,

$\int_{|z|=3} f(z)dz = 10 \pi i$

Report an Error

Example Question #6 : 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

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

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

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

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 |z|=3 .

Possible Answers:

$- \pi i$

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

$2 \pi i$

$4 \pi i$

Correct answer:

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{1}{1+\frac{1}{z^3}} = \frac{1}{z^2+z^{-1}} = \frac{ z}{ 1 + z^3 }$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

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

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

Thus,

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

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,

$\int_{|z|=2} f(z)dz =0$

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Example Question #1 : 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

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

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

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

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 |z|=4 .

Possible Answers:

$\pi i$

$- 2 \pi i$

$4 \pi i$

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

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

Correct answer:

$4 \pi i$

Explanation:

Note,

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

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

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

Thus,

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

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,

$\int_{|z|=2} f(z)dz =4 \pi i$

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Example Question #1 : 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

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

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

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

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 |z|=3 .

Possible Answers:

$\8 \pi i$

$4 \pi i$

$2 \pi i$

$14 \pi i$

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

Correct answer:

$14 \pi i$

Explanation:

Note,

$\frac{1}{z^2}f(\frac{1}{z}) = \frac{1}{z^2} \frac{7 \frac{1}{z} - 3}{ \frac{1}{z} (\frac{1}{z} - 1)} = \frac{7 \frac{1}{z} -3 }{ 1 -z} = \frac{7}{z (1-z)} - \frac{3}{1 -z}$

Thus, seeking to apply the Residue Theorem above for $\frac{1}{z^2}f(\frac{1}{z})$ inside |z|=3 , we evaluate the residue for z=0 .

Observe,

$\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{7}{z (1-z)} - \frac{3}{1 -z} \newline = \frac{7}{z}\displaystyle \sum_{k=0}^{\infty} z^{k} - 3\displaystyle \sum_{k=0}^{\infty} z^{k} \newline = 7 \displaystyle \sum_{k=0}^{\infty} z^{k-1} - 3 \displaystyle \sum_{k=0}^{\infty} z^{k}$

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

Thus,

$\text{Res}_{z=0} [\frac{1}{z^2}f(\frac{1}{z})] = 7$ .

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,

$\int_{|z|=2} f(z)dz =14 \pi i$

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Example Question #9 : Residue Theory

Find the residue of the function

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

Possible Answers:

$\frac{7}{9}$

$- \frac{2}{945}$

$-\frac{1}{45}$

$\frac{1}{2}$

$- \frac{2}{3}$

Correct answer:

$- \frac{2}{945}$

Explanation:

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,

$\text{Res}_{z=0}[f(z)] = \text{Res}_{z=0}[\frac{cot(z)}{z^6}] = - \frac{2}{945}$ .

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Example Question #51 : Complex Analysis

Find the residue at z=0 of

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

Possible Answers:

$\pi$

$\frac{1}{4}$

Correct answer:

Explanation:

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,

$\text{Res}_{z=0}[f(z))] = \text{Res}_{z=0}[z \cos (\frac{1}{z})] = 0$

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Example Question #51 : Complex Analysis

Find the residue at z = 0 for the function

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

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

$\pi$

Correct answer:

Explanation:

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 4*1=4 .

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0} [\frac{4}{z+z^2}] = 4$ .

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Example Question #51 : Complex Analysis

Find the residue at z=0 of the function

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

Possible Answers:

$\frac{7}{9}$

$-\frac{1}{2}$

Correct answer:

Explanation:

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 $\frac{(-1)^0}{(2*0+1)!}= 1$ .

Thus,

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

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Example Question #1 : Applications Of Harmonic Functions

If f(z) = u(x,y) + iv(x,y) and $\overline{f(z)} = u(x,y) - iv(x,y)$ are both analytic through a domain , which of the following is true?

Possible Answers:

f(z) = 0

f(z) must be constant throughout

f(z) is nowhere analytic

f(z) is nowhere differentiable

None of the other answers

Correct answer:

f(z) must be constant throughout

Explanation:

Since f(z) = u(x,y) + iv(x,y) is analytic, the Cauchy-Riemann Equations give us $u_x(x,y) = v_y(x,y) \text{ and } u_y(x,y) = -v_x(x,y) \forall x,y$

Since $\overline{f(z)} = u(x,y) - iv(x,y)$ is analytic, the Cauchy-Riemann Equations give us $u_x(x,y) = -v_y(x,y) \text{ and } u_y(x,y) = v_x(x,y) \forall x,y$ .

Putting these two results together gives us $u_x(x,y) = 0 \text{ and } u_y(x,y) = 0 \forall x,y$

which implies that u(x,y) is constant. Similar reasoning gives is that v(x,y) is constant. This implies that f(z) must be constant throughout .

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Complex Analysis : Complex Analysis

Study concepts, example questions & explanations for Complex Analysis

All Complex Analysis Resources

Example Questions

Example Question #4 : Residue Theory

Example Question #1 : Residue Theory

Example Question #6 : Residue Theory

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #9 : Residue Theory

Example Question #51 : Complex Analysis

Example Question #51 : Complex Analysis

Example Question #51 : Complex Analysis

Example Question #1 : Applications Of Harmonic Functions

All Complex Analysis Resources

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