I started complex analysis some weeks back and now im doing a review and i came across this integral! The problem in question is already solved in my exercise session but there were some steps which i did not understand. Ok this is how the problem goes
Compute the integral given that $A=\{z\in C: \frac 12 \leq |z| \leq2, |Im(z)| \le |Re(z)|\}$
$$\int_{\partial{A}}{\frac 1{(z^2-z)^2}dz}$$
So first you find the singularities right? $z_0$=0 and $z_1$=1
Both of them are poles of the order 2
$z_0$=0 $\notin$ A so therefore we don't even consider it?? Correct me if I'm wrong
So the residue of $z_1=1$ is $$Res(f_1,1)=???$$ Here he directly jumps to the computation of the limit. Here is what he does.
$$ Res(f_1,1)=\lim_{z\to1} \frac d{dz} \left[\frac 1{z^2}\right] \ldots=-2\implies \int_{\partial{A}}{\frac 1{(z^2-z)^2}dz}=-4\pi i $$
What i dont get is How did he reach to the limit part? Can someone help me?
Let's consider a function which has a pole of order 2 at $z_{0}$. It is in the form: $$f(z)=\frac{a_{-2}}{(z-z_{0})^2}+\frac{a_{-1}}{(z-z_{0})}+a_0+a_{1}(z-z_0)....$$
Our objective is to find the residue at $z_{0}$ (ie) $a_{-1}$. So we multiply both sides by $(z-z_0)^2$ to separate out $a_{-1}$
$$(z-z_0)^2f(z)=a_{-2}+a_{-1}(z-z_0)+a_{0}(z-z_0)^2....$$ Differentiating with respect to $z$ on both sides: $$\frac{d}{dz}[(z-z_0)^2f(z)]=a_{-1} + 2a_{0}(z-z_{0})...$$ So now, if $z\to z_0$, all terms except $a_{-1}$ becomes $0$. $$a_{-1}=\lim_{z\to z_0}\frac{d}{dz}[(z-z_0)^2f(z)]=Res(f,z_{0})$$ So in your case $Res(f,1)$ is: $$\lim_{z\to 1}\frac{d}{dz}(z-1)^2\frac{1}{z^2\cdot(z-1)^2}=\lim_{z\to 1}\frac{d}{dz}\frac{1}{z^2}=-2$$