Integration of $1/|z-z_0|^2$ over a circle in the complex plane I am trying to prove the following problem
$$\frac{1}{2\pi R}\int_{\delta B_R(0)}\frac{|dz|}{|z-z_0|^2}=\frac{1}{|R^2-|z_0|^2|}$$ where $\delta B_R(0)$ is the boundary of a circle with centre (0,0) and radius R.
Now I have used $z=Re^{i\theta}$ and got the following $$\frac{1}{2\pi i}\int_{\delta B_R(0)}\frac{dz}{z|z-z_0|^2}$$ after that I am stuck. Any help would be highly appreciated. (Someone told me that I need to use Poisson kerenl, but I have no idea how to use that here)
 A: Denote by $z=Re^{i\theta}$ and $z_0=re^{i\theta_0}$
$$\frac{1}{2\pi R}\int_{\delta B_R(0)}\frac{|dz|}{|z-z_0|^2}=\frac{1}{2\pi R}\int^{2\pi}_{0}\frac{|Rie^{i\theta}\,d\theta|}{|Re^{i\theta}-re^{i\theta_0}|^2}=\frac{1}{2\pi}\int^{2\pi}_{0}\frac{d\theta}{|Re^{i\theta}-re^{i\theta_0}|^2}=\frac{1}{2\pi}\int^{2\pi}_{0}\frac{d\theta}{(Re^{i\theta}-re^{i\theta_0})(Re^{-i\theta}-re^{-i\theta_0})}=\frac{1}{2\pi}\Big(\int^{2\pi}_{0}\frac{d\theta}{(R^2-2Rr\cos(\theta-\theta_0)+r^2)}\Big)=\frac{1}{2\pi}\Big(\int^{2\pi-\theta_0}_{-\theta_0}\frac{d\alpha}{(R^2-2Rr\cos(\alpha)+r^2)}\Big)$$
In the last step we did the substitution $\alpha=\theta_0-\theta$.
Denote by $c=\frac{R^2+r^2}{2Rr}$ (note that $c\geq 1$). We can express the last integral as
$$\frac{1}{4\pi Rr}\Big(\int^{2\pi-\theta_0}_{-\theta_0}\frac{d\alpha}{c-\cos(\alpha)}\Big)$$
Notice that $$\cos(\alpha)=\frac{e^{i\alpha}+e^{-i\alpha}}{2}=\frac{w+1/w}{2}$$
So $d\alpha=dw/iw$. One is able to express the last integral in terms of $w$ as follows
$$\frac{1}{4\pi Rr}\Big(\oint_{|w|=1}\frac{1}{c-\frac{w+1/w}{2}}\frac{dw}{iw}\Big)=\frac{1}{2\pi i Rr}\Big(\oint_{|w|=1}\frac{1}{-w^2+2cw-1}\,dw\Big)=-\frac{1}{2\pi i Rr}\Big(\oint_{|w|=1}\frac{1}{w^2-2cw+1}\,dw\Big)$$
The integrand has two poles 
$c+\sqrt{c^2-1}$ and $c-\sqrt{c^2-1}$. Only one root lies within the unit circle namely $$w_0=c-\sqrt{c^2-1}$$ with residue $-\frac{1}{2\sqrt{c^2-1}}$
Therefore by the residue theorem we can obtain the desired integral as 
$$-\frac{1}{2\pi i Rr}\Big(\oint_{|w|=1}\frac{1}{w^2-2cw+1}\,dw\Big)=-\frac{1}{2\pi i Rr}\cdot 2\pi i\cdot(-\frac{1}{2\sqrt{c^2-1}})=\frac{1}{2Rr\sqrt{c^2-1}}$$
Substitute back $c=\frac{R^2+r^2}{2Rr}$ to obtain the final result
$$\frac{1}{2Rr\sqrt{c^2-1}}=\frac{1}{\sqrt{(R^2+r^2)^2-4R^2r^2}}=\frac{1}{|R^2-r^2|}$$
And you are done.
