1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex integration (Using Cauchy Integral formula)

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data

    $$\int_\gamma \frac{\cosh z}{2 \ln 2-z} dz$$

    with ##\gamma## defined as:

    1. ##|z|=1##
    2. ##|z|=2##

    I need to solve this using Cauchy integral formula.

    2. Relevant equations

    Cauchy Integral Formula

    3. The attempt at a solution

    With ##|z|=2## I've solved already, as it is quite easy. All one needs to do is re-write the denominator as ##z-2 \ln 2## and since ##2 \ln 2## is inside ##\gamma##, then one can apply easily the formula.

    With 1 I'm not sure how to re-write the integral so I can use the Cauchy Integral formula.
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 19, 2013 #2
    This has been solved already. No need to answer.
     
  4. Oct 19, 2013 #3
    For others who are still interested: obviously the function is holomorphci in the circle with radius 1.
     
  5. Oct 20, 2013 #4
    Oh, right.

    So yes, by the Cauchy theorem, it's 0.

    One could always define ##f(z)## as

    $$f(z)=\frac {z \cosh z}{2 \ln 2 -z}$$

    and so the Integral becomes

    $$\int_\gamma \frac {f(z)}{z}=0$$
     
    Last edited: Oct 20, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complex integration (Using Cauchy Integral formula)
Loading...