Let $a \in \mathbb{C}$ with $|a|>1$. I need to evaluate the path integral around the unit circle in $\mathbb{C}$: $$\int_{|z|=1}\frac{|dz|}{|az-1|^2}$$ where $|dz|$ represents integration with respect to arc-length.
My solution is following:
Since $z$ is the unit circle, we can rewrite it as $z=x+iy=e^{it}$ with $x=\cos t$ and $y=\sin t$. Hence, $|dz|=dt$ after the required transformations. Also, if $a=re^{i\varphi} (r>1)$, we have: $$\int_{|z|=1}\frac{|dz|}{|az-1|^2}=\int_{0}^{2\pi}\frac{dt}{|re^{i(t+\varphi)}-1|^2}=\int_{0}^{2\pi}\frac{dt}{(r\cos (t+\varphi)-1)^2+r^2\sin (t+\varphi)^2}=\int_{0}^{2\pi}\frac{dt}{r^2+1-2r\cos (t+\varphi)}$$ and then I can try to find out the integral, but it is a little bit challenging.
So, my question is if I did everything right? and are there any easier ways to calculate this?
Any hints are welcome! Thank you!
To treat this integral as a contour integral along the unit circle, the appropriate substitution is $$ |dz| = dt = \frac{dz}{i z} $$ on recalling that $z = e^{i t}, dz = i e^{i t} dt$ and $|dz| = dt$.
As for the integrand itself, $$ |az - 1|^2 = (\bar a \bar z - 1)( a z - 1) = \bigg ( \frac{\bar a}{z} - 1\bigg) (a z - 1) \, . $$ This uses the fact that $\bar z = 1/z$ for $|z| = 1$. Collecting, $$ \int \frac{|dz|}{|a z - 1|^2} = \int \frac{dz}{i (\bar a - z)(a z - 1)} \, , $$ which can now be solved using residue calculus to resolve the pole at $z = 1/a$.