Daily Chaos: Cauchy's Integral Formula

Thursday, December 17, 2015

f $\gamma$ is a simple closed contour and if $f(z)$ is analystic on and inside of $\gamma$ then

$\oint_{\gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz = \frac{2\pi i}{n!}f^{n}(z_0)$

Evaluate the following four integrals

$\oint_{|z|=3} \frac{e^{iz}}{z+i}dz =2\pi i e$

$\oint_{|z|=3} \frac{\sin(2\pi z)}{3z-1}dz =\frac{\sqrt{3}\pi i}{3}$

$\oint_{|z|=3} \frac{z+3}{(z-2)(z+4)^2}dz = \frac{2\pi i}{9}$

$\oint_{|z|=3} \frac{e^{iz}}{(z+1)^2}dz =-2\pi e$