<s>[INST] Residues of poles Find $Res_{f}\left ( z_{0} \right )$, where, $f\left ( z \right )=\frac{1}{z^{4}+4}$, for $z_{0}=1+i$
The definition for 
$$Res_{f}\left ( 1+i \right ) =\lim_{z \to z_{0}} \left\{\left ( z-\left ( 1+i \right ) \right ) \cdot \frac{1}{z^{4}+4} \right\}$$
and the roots for
$$z^{4}+4$$ are $\sqrt{+2i}$, $\sqrt{-2i}$, $- \sqrt{+2i}$, $- \sqrt{-2i}$
I'm a bit stuck here. Could someone give me a push?
 [/INST] Using $1 + i = \sqrt{2} \, e^{\tan^{-1}(1)} = \sqrt{2} \, e^{\pi i/4}$ and $\sqrt{i} = e^{\pi \, i/4}$ then
\begin{align}
\lim_{z = 1+i} f &= \lim_{1+i} \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{z^{4}+4} \right\} \\
&= \lim \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{(z^{2} + 2i)(z^{2}- 2 i) } \right\} \\
&= \lim \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{(z^{2} + 2 i) (z - \sqrt{2} e^{\pi i/4})( z + \sqrt{2} e^{\pi i/4})} \right\} \\
&= \frac{1}{(2 \, e^{\pi i/2} + 2 i)(2 \, \sqrt{2} \, e^{\pi i/4}) } \\
&= \frac{1}{8 \sqrt{2}} \, \frac{1}{ e^{3 \pi i/4}} \\
&= - \frac{1 + i}{16}
\end{align}
</s>