Let $f(z)=z^5-3iz^2+2z-1+i$. Evaluate the integral of $\frac{f'(z)}{f(z)}$ around a contour $C$ where $C$ encloses all the zeroes of $f$.
I'm not sure what to do here. It seems unlikely I should be expected to find the roots of $f$ so as to use the residue theorem
This is a more detailed hint than what I provided in comments, since I just confirmed the direction I had in mind actually works.
Given $z_0$ (some root of $f$ with multiplicity $m$) write $f(z) = (z-z_0)^m g(z)$, for some polynomial $g$ which is nonzero at $z_0$.
This should aid you in first prooving that all the poles of $\frac{f'(z)}{f(z)}$ are simple, then evaluating the limit you will then know is the residue at $z_0$ - happily independent of the actual value of the root.
Edit: A large, and hopefully helpful prod in the right direction.
Note that: $$f'(z) = m(z-z_0)^{m-1} g(z)+(z-z_0)^m g'(z) = (z-z_0)^{m-1} (m g(z)+(z-z_0)g'(z))$$
Thus $$\frac{f'(z)}{f(z)} = \frac{(z-z_0)^{m-1} (m g(z)+(z-z_0)g'(z))}{(z-z_0)^m g(z)} = \frac{m g(z)+(z-z_0)g'(z)}{(z-z_0) g(z)} = \frac{m}{z-z_0}+\frac{g'(z)}{g(z)}$$
Perhaps you can take it from here.