Use Cauchy's Integral Formula to evaluate the following integral:
$$\int\limits_\Gamma \frac{1}{{(z-1)^3}{(z-2)^2}}dz$$ where $$\Gamma$$is a circumference of radius $4$ centered at $-2+i$ and traversed once in the positive(with respect to the interior of the disk) direction.
My thoughts on the problem:
I HAVE to use the Cauchy Integral Formula. I've been trying to decide the best way to change the expression in the integral. If I change it to:
$$\int\limits_\Gamma \frac{\frac{1}{{(z-2)^2}}}{{(z-1)^3}}$$
The point 2 is on the boundary of Gamma which means I can NOT use the formula. Are there any other ideas of ways I could change this integral to make it friendly enough to use the formula?
Best Answer
Be careful, $2$ is not on the boundary of $\Gamma$. Then your approach is the correct one, letting $f(z)=1/(z-2)^2$, then by the General Cauchy Integral formula: $$ \int_{\Gamma} \frac{f(z)}{(z-z_0)^3} = 2\pi i \frac{f''(z_0)}{2!} = \frac{6 \pi i}{(z_0-2)^4} = 6\pi i $$ since $f''(z)=6/(z-2)^4$ and $z_0=1$.