a contour integration involving $5$th roots of unity 
Let $\gamma$ be the circle of radius $2$ with centre at the origin in
  the complex plane oriented in the anti-clockwise direction, then the
  integral $$\oint_\gamma \frac{dz}{(z-3)(z^5-1)}$$ equals
$(a)$ $\displaystyle{\frac{2\pi i}{3^5-1}}$.
$(b)$ $\displaystyle{\frac{2\pi i}{3^4-1}}$.
$(c)$ $\displaystyle{-\frac{2\pi i}{3^5-1}}$.
$(d)$ $0$.

Clearly by Rouche's theorem, all the zeroes of $z^5-1$ lies on the circle $|z|=1$. So if we count those zeroes by $z_k$ where $k=0,1,2,3,4$ we have, by Residue theorem $$\oint_\gamma \frac{dz}{(z-3)(z^5-1)}=-\frac{2\pi i}{5}\sum_{k=0}^4 \frac{1}{3{z_k}^4-1}$$ which is a nightmare to simplify algebraically. Can someone help in this regard if I have done something wrong? Thanks in advance.

Hint. By using the residue at infinity we have that
$$\oint_{|z|=2} \frac{dz}{(z-3)(z^5-1)}=-2\pi i(\text{Res}(f,3)+\text{Res}(f,\infty)).
$$
and those two residues are quite easy to evaluate for $f(z)=\frac{1}{(z-3)(z^5-1)}$.
Recall that on the Riemann sphere, the algebraic sum of the complex residues of a meromorphic function is zero. See also the Inside-Outside Theorem.
