Let $a\in \mathbb C, r>0$ and $\gamma_r=\partial D(0,r)$. I want to evaluate the following line integral
$I=\int_{\gamma_r}\frac{1}{|z-a|^2}ds.$
I'm looking for a complex function $g(z)$ such that
$I=\int_{\gamma_r}g(z)dz.$
I put $g(z)=\dfrac{1}{iz|z-a|^2}$. Then
$\int_{\gamma_r}g(z)dz=\int_0^{2\pi}\frac{rie^{it}}{rie^{it}|re^{it}a|^2}dt=\frac{1}{r}I.$
If $f(z)=\dfrac{1}{i|z-a|^2}$, I can apply Cauchy's integral formula to conclude
$f(0)=\frac{1}{2\pi i}\int_{\gamma_r}\frac{f(z)dz}{z}dz=\frac{1}{2\pi i}\int_{\gamma_r}g(z)dz.$
I find
$I=r\int g(z)dz=2\pi r i f(0)=\frac{2\pi r}{|a|^2}.$
Unfortunately, the correct solution should be
$\dfrac{2\pi r}{| |a|^2-r^2|}.$
Where did I go wrong?