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Homework Help: Cauchy's Theorem and Cauchy's Formula Exercises

  1. Jun 27, 2013 #1
    I am studying for a test I have tomorrow evening and have finally reached the final section. I'm not 100% whether or not I am approaching these questions correctly or not so please bear with me. I have four questions and am not sure whether it merits starting separate threads considering they are similar - therefore, I will attempt all here and will only start separate threads if requested. Thanks a bunch in advance for any and all assistance!

    1. The problem statement, all variables and given/known data
    "In Exercises 1 to 4, evaluate the given integral using Cauchy's Formula or Theorem.

    1. ##\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz##
    2. ##\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz##
    3. ##\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz##
    4. ##\int_{|z|=1} \frac{sin(z)}{z} dz##" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 116)

    2. Relevant equations
    "Cauchy's Theorem Suppose that f is analytic on a domain D. Let ##\gamma## be a piecewise smooth simple closed curve in D whose inside Ωalso lies in D. Then

    $$\int_{\gamma} f(z) dz = 0$$" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 106)

    "Cauchy's Formula Suppose that f is analytic on a domain D and that ##\gamma## is a piecewise smooth, positively oriented simple closed curve in D whose inside Ω also lies in D. Then

    $$f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(a)}{(a - z)} da$$ for all ##z \in## Ω" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 111)

    3. The attempt at a solution

    Exercise 1:

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

    For this exercise, ##\gamma = |z| = 1## and so this integral cannot be rewritten in the form required by Cauchy's Formula. We use Cauchy's Theorem instead. Therefore,

    $$\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz = 0$$

    Exercise 2:

    Here ##\gamma = |z| = 2## and so we can write this integral in the form of Cauchy's Formula:

    $$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz$$
    $$\int_{|z| = 2} \frac{e^{z}}{(z-0)(z-3)} dz$$
    $$\int_{|z| = 2} \frac{g(z)}{(z-0)} dz, g(z) = \frac{e^{z}}{z-3}$$
    $$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz = (2\pi i)g(0) = (2\pi i)\frac{e^{0}}{0-3} = (2\pi i)\frac{-1}{3}$$

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

    Exercise 3:

    This exercise has ##\gamma = |z+1| = 2##.

    $$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz$$
    $$\int_{|z+1| = 2} \frac{z^{2}}{(2-z)(2+z)} dz$$
    $$\int_{|z+1| = 2} \frac{z^{2}}{-(z-2)(z+2)} dz$$

    We find ##g(z) = \frac{z^{2}}{-(z-2)} = \frac{z^{2}}{2-z}## and evaluate ##g(z)## at ##z = -2)##:

    $$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz = (2\pi i)g(-2) = (2\pi i)\frac{(-2)^{2}}{2-(-2)} = (2\pi i)\frac{4}{4} = 2\pi i$$

    Exercise 4:

    Here we have ##\gamma = |z| = 1## and:

    $$\int_{|z|=1} \frac{sin(z)}{z} dz$$
    $$\int_{|z|=1} \frac{sin(z)}{z-0} dz$$

    We say ##g(z) = sin(z)## so we can write:

    $$\int_{|z|=1} \frac{sin(z)}{z-0} dz = (2\pi i)g(0) = (2\pi i)sin(0) = 0$$

    Is this the proper approach to solving these four problems? I get the same final answer for Exercises 1 and 3, but the answers to 2 and 4 are not provided. Thanks very much for your time! :)
  2. jcsd
  3. Jun 27, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    I only checked 2 and 4 and skimmed over 1 and 3, but you seem to have a pretty good idea of what you are doing and those are correct. Well done!
  4. Jun 27, 2013 #3
    Thank you very much for the prompt reply. It's difficult to be sure I understand when there are only four exercises (directly) related to this material in the book.
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