1. ## Evaluate Complex Integral

Use Cauchy's Integral Formula to evaluate the integral $\oint_{C}f(z)dz$ over a contour C, where C is the boundary of a square with diagonal opposite corners at z = −(1 + i )R and z = (1 + i )R, where R > a > 0, and where $f(z)=\frac{e^z}{z-\frac{i\pi}{4} a}$

I know f(z) would not be analytic at $\frac{i\pi}{4} a$.

Then by Cauchy's Integral Formula:

$f(a) = \frac{1}{2i\pi} \oint_{C}\frac{f(z)}{z-a}$

$f(\frac{i\pi}{4} a) = \frac{1}{2i\pi} \oint_{C}\frac{e^z}{z-\frac{i\pi}{4} a}$

Am I going in the right direction so far?

2. ## Re: Evaluate Complex Integral

Originally Posted by ivinew
Use Cauchy's Integral Formula to evaluate the integral $\oint_{C}f(z)dz$ over a contour C, where C is the boundary of a square with diagonal opposite corners at z = −(1 + i )R and z = (1 + i )R, where R > a > 0, and where $f(z)=\frac{e^z}{z-\frac{i\pi}{4} a}$

I know f(z) would not be analytic at $\frac{i\pi}{4} a$.

Then by Cauchy's Integral Formula:

$f(a) = \frac{1}{2i\pi} \oint_{C}\frac{f(z)}{z-a}$

$f(\frac{i\pi}{4} a) = \frac{1}{2i\pi} \oint_{C}\frac{e^z}{z-\frac{i\pi}{4} a}$

Am I going in the right direction so far?
Well the symbol $f$ is already "taken," so basically you want to let $g(z)=e^z$ and notice that

$e^{i\pi a/4}=g(i\pi a/4)=\frac{1}{2\pi i}\int_C\frac{g(z)}{z-(i\pi a/4)} dz$,

and you're done.