We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Complex Analysis : Residue Theory

Study concepts, example questions & explanations for Complex Analysis

Create An Account

All Complex Analysis Resources

1 Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

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

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 .

Possible Answers:

$- 2 \pi i$

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

$3 \pi i$

$\pi i$

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

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,

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

Report an Error

Example Question #2 : 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

Possible Answers:

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

$-e^{2 \pi i}$

$2 \pi i$

$- 2 \pi i$

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

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,

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

Report an Error

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

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

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 |z|=4 .

Possible Answers:

$- \frac{\pi i }{6}$

$\frac{\pi i}{12}$

$2 \pi i$

$\frac{\pi i }{24}$

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

Correct answer:

$\frac{\pi i}{12}$

Explanation:

Note, there is one singularity for f(z) where z = 0 .

Let

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

Then

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

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

Therefore, there is one singularity for f(z) where u = 0 . Hence, we seek to compute the residue for f(z) where u = 0 = z

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 k = 4 , $z^{3-k} = z^{-1} = \frac{1}{z}$ .

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

Therefore,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0} [z^2 e^{\frac{1}{z}}] = \frac{1}{4!}$

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,

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

Report an Error

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

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

in the region |z|=3 .

Possible Answers:

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

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

$\pi i$

$2 \pi i$

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,

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

Report an Error

Example Question #2 : 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 #3 : 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:

$- \pi i$

$2 \pi i$

$4 \pi i$

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

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,

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

Report an Error

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

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

in the region |z|=4 .

Possible Answers:

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

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

$\pi i$

$4 \pi i$

$- 2 \pi i$

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,

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

Report an Error

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:

$\8 \pi i$

$4 \pi i$

$14 \pi i$

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

$2 \pi i$

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,

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

Report an Error

Example Question #1 : Residue Theory

Find the residue of the function

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

Possible Answers:

$\frac{1}{2}$

$- \frac{2}{945}$

$-\frac{1}{45}$

$\frac{7}{9}$

$- \frac{2}{3}$

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,

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

Report an Error

Example Question #2 : Residue Theory

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,

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

Report an Error

View Tutors

Hannah
Certified Tutor

Columbia International University, Bachelor of Science, Humanities.

View Tutors

Enock
Certified Tutor

University of Bridgeport, Current Undergrad Student, Computer Engineering, General.

View Tutors

Tracy
Certified Tutor

University of West Georgia, Doctor of Education, Instructional Technology.

All Complex Analysis Resources

1 Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Physics Tutors in Atlanta, Reading Tutors in Washington DC, French Tutors in Phoenix, Math Tutors in Dallas Fort Worth, ISEE Tutors in Houston, Statistics Tutors in Denver, Chemistry Tutors in Los Angeles, LSAT Tutors in Chicago, Math Tutors in Los Angeles, Statistics Tutors in Boston

Popular Courses & Classes

ACT Courses & Classes in Philadelphia, GRE Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in New York City, ISEE Courses & Classes in Miami, SSAT Courses & Classes in Miami, SAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Phoenix, MCAT Courses & Classes in Denver, ACT Courses & Classes in Atlanta, SAT Courses & Classes in Washington DC

Popular Test Prep

ACT Test Prep in Atlanta, MCAT Test Prep in Seattle, MCAT Test Prep in Denver, LSAT Test Prep in Seattle, GRE Test Prep in New York City, SSAT Test Prep in San Diego, ISEE Test Prep in New York City, GRE Test Prep in Washington DC, ACT Test Prep in Philadelphia, GRE Test Prep in Chicago

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Complex Analysis : Residue Theory

Study concepts, example questions & explanations for Complex Analysis

All Complex Analysis Resources

Example Questions

Example Question #1 : Residue Theory

Example Question #2 : Residue Theory

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #2 : Residue Theory

Example Question #3 : Residue Theory

Example Question #4 : Residue Theory

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #2 : Residue Theory

All Complex Analysis Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: