(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Use Cauchy's Integral Formula to compute the following integral:

[tex]\int_{|z|=2\pi}\frac{\pi \sin(z)}{z(z-\pi)}dz[/tex]

2. Relevant equations

[tex]\int_C \frac{f(z)dz}{z-z_0}=2\pi if(z_0)[/tex]

3. The attempt at a solution

If we let

[tex]f(z)=\frac{\pi \sin(z)}{z}[/tex]

and

[tex]z_0=\pi[/tex]

Then

[tex]\int_{|z|=2\pi}\frac{\pi \sin(z)/z}{(z-\pi)}dz=0[/tex]

Is this correct? I'm concerned about the fact that the function isn't differentiable at z=pi and that pi is an interior point of C...Part of me thinks that this integral cannot be evaluated with Cauchy's formula, but I do not yet understand the subject matter well enough to know for sure.

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# Homework Help: Cauchy's Integral Formula Problem

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