$\displaystyle \lim_{z\rightarrow -i}\frac{z^4-1}{z+ i}$
What do I do with a Complex limit when I get $\displaystyle \frac{0}{0}\mbox{?}$
You can also use L'Hospital's Rule.
$\displaystyle \displaystyle \lim_{z \to -i}\frac{z^4 - 1}{z + 1} = \lim_{z \to -i}\frac{\frac{d}{dz}(z^4 - 1)}{\frac{d}{dz}(z + 1)}$
$\displaystyle \displaystyle = \lim_{z \to -i}\frac{4z^3}{1}$
$\displaystyle \displaystyle = \lim_{z \to -i}{4z^3}$
$\displaystyle \displaystyle = 4(-i)^3$
$\displaystyle \displaystyle = 4i$.
This is also what you would get if you had factorised and cancelled the common factor of $\displaystyle \displaystyle z+i$.