How do I solve this complex integration problem?. I want to find the value of  $$I=\int_{|z|=r}\frac{|dz|}{|z-z_0|^4},$$ where $|z_0|\neq r>0$.
 A: On the integration contour, $z \bar{z} = r^2$. Also, $z=r e^{i \phi} \implies dz = i r e^{i \phi} d\phi$, or $|dz| = -i r \, dz/z$  Thus, $|dz|=r d\phi$.  Further,
$$|z-z_0|^4 = (|z-z_0|^2)^2 = \left (r^2+z_0^2 - \bar{z_0} z - \frac{r^2}{z} z_0 \right )^2 $$
Thus, the integral is equal to
$$-i r \oint_{|z|=r} dz \frac{z}{\left (\bar{z_0} z^2  - (r^2+|z_0|^2) z  + r^2 z_0 \right )^2} $$
This may be evaluated using the residue theorem.  The roots of the denominator, i.e., the poles, are at 
$$z_+ = z_0 \quad z_- = \frac{r^2}{\bar{z_0}} $$
Assume $|z_0| \lt r$.  Then the only pole we need to deal with is $z=z_+=z_0$ because $|z_-| \gt r$ in this case.  Then the integral is equal to, when $|z_0| \lt r$,
$$-i \frac{r}{\bar{z_0}^2} \oint_{|z|=r} dz  \frac{z}{(z-z_0)^2 \left (z-\frac{r^2}{\bar{z_0}} \right )^2} = 2 \pi \frac{r}{\bar{z_0}^2} \left [\frac{d}{dz} \frac{z}{\left (z-\frac{r^2}{\bar{z_0}} \right )^2} \right ]_{z=z_0}$$
And you should be able to take it from here, including the case where $|z_0| \gt r$.  The end result is
$$\oint_{z=r} \frac{|dz|}{|z-z_0|^4} = 2 \pi r \frac{r^2+|z_0|^2}{\left |r^2-|z_0|^2 \right |^3} $$
