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?
 A: 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}
A: With $\rho$ a simple zero of $f(z)$ we have
$$\mathrm{Res}_{z=\rho} \frac{1}{f(z)} = \frac{1}{f'(\rho)}.$$
This gives for the present case
$$\frac{1}{4\rho^3} = \frac{\rho}{4\rho^4}
= \frac{\rho}{4(-4)} = -\frac{1+i}{16}.$$
Consult e.g. Wikipedia for more information.
A: Note: This is only a supplement to Leucippus nice answer, which may assist OPs understanding

In order to find 
\begin{align*}
\mathop{Res}_f(1+i)=\lim_{z\rightarrow 1+i}\frac{z-(1+i)}{z^4+4}
\end{align*}
  we obtain some other valid representations of $z_0=1+i$.

First note, that $$(1+i)^2=1+2i-1=2i$$ which implies $1+i=\sqrt{2i}$.
On the other hand we can write $1+i$ using polar coordinates and find
$$1+i=\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)=\sqrt{2}e^{i\frac{\pi}{4}}$$

So, we can alternatively use
  \begin{align*}
1+i=\sqrt{2i}=e^{i\frac{\pi}{4}}
\end{align*}

Now, we can calculate the residuum at $z_0=1+i$

\begin{align*}
\mathop{Res}_f(1+i)&=\lim_{z\rightarrow 1+i}\frac{z-(1+i)}{z^4+4}\\
&=\lim_{z\rightarrow 1+i}\frac{z-(1+i)}{(z^2+2i)(z^2-2i)}\\
&=\lim_{z\rightarrow \sqrt{2i}}\frac{z-\sqrt{2i}}{(z^2+2i)(z-\sqrt{2i})(z+\sqrt{2i})}\tag{1}\\
&=\lim_{z\rightarrow \sqrt{2i}}\frac{1}{(z^2+2i)(z+\sqrt{2i})}\\
&=\frac{1}{(4i)(2\sqrt{2i})}\\
&=\frac{\sqrt{2i}}{-16}\\
&=-\frac{1+i}{16}\\
\end{align*}

Comment: In (1) we use the representation $1+i=\sqrt{2i}$
