contour integral with arclength Let |c|< 1 and $\gamma$ be the unit circle , then how do i calculate the following integral?
$$\oint_\gamma \frac{|dz|} {{|z-c|^2}}$$
I've tried writing the integral with  $d\theta$ but this led nowhere with the absolute value
in the denominator.Any hints?
 A: A hint for doing this with residue calculus: recall that for $z$ on the unit circle, we have that $z \overline{z} = 1$. This allows us to get the conjugate $\overline{z}$ out of the expression and to make the integrand holomorphic.
Also, you might want to show that when the integral is over the unit circle,
$$
|dz| = \frac{dz}{i z}
$$
A: Let $c=re^{it}$. The integral is now
$$
\int_0^{2\pi}\frac{d\theta}{|e^{-i\theta}-re^{it}|^2}.
$$
We see that
$$
|a-b|^2=\overline{(a-b)}(a-b)=\overline{a}a-\overline{a}b-\overline{b}a+\overline{b}b=|a|^2-2\rm{Re}(a-b)+|b|^2
$$
for all $a,b\in\mathbb{C}$. By substituting $a=e^{i\theta}$ and $b=re^{it}$ to this, we get
$$
|e^{-i\theta}-re^{it}|^2=1-2\rm{Re}(re^{i(t-\theta)})+r^2=1-2r\cos(t-\theta)+r^2
$$
so the integral is
$$
\int_0^{2\pi}\frac{d\theta}{1-2r\cos(t-\theta)+r^2}=\frac{2\pi}{1-r^2}\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(t-\theta)+r^2}d\theta.
$$
We see the Poisson kernel
$$
P_r(t-\theta)=\frac{1-r^2}{1-2r\cos(t-\theta)+r^2}
$$
in our last integral. Since
$$
\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(t-\theta)+r^2}d\theta=1,
$$
the integral is
$$
\int_0^{2\pi}\frac{d\theta}{|e^{-i\theta}-re^{it}|^2}=\frac{2\pi}{1-r^2}=\frac{2\pi}{1-|c|^2}.
$$
