Here is the second one. We need to use the residue theorem here. Look at picture below. The red dots are the poles.
Theorem: Let $\displaystyle n$ be a non-negative integer then $\displaystyle \mbox{res}(\Gamma(z),-n) = \frac{(-1)^n}{n!}$.
Thus, by the residue theorem we have,
$\displaystyle 2\pi i \left( \frac{1}{2!} - \frac{1}{1!}+\frac{1}{0!} \right) = \pi i$
The first is:
$\displaystyle \frac{\Gamma(2+2i)}{\Gamma(-1+2i)}$
Now to do this you need a basic recurrence for $\displaystyle \Gamma(z)$, this is:
$\displaystyle \Gamma(1+z)=z\Gamma(z)$.
Using this we have:
$\displaystyle \Gamma(2+2i)= (1+2i)(2i)(-1+2i) \Gamma(-1+2i)$,
so:
$\displaystyle
\frac{\Gamma(2+2i)}{\Gamma(-1+2i)}= (1+2i)(2i)(-1+2i)
$
RonL