Find all the singularities of $f(z)= \frac{1}{z^4+1}$ and the associated residue for each singularity I know that there are poles at
$$\Large{z=e^{\frac{i\pi}{4}}},$$
$$\Large{z=-e^{\frac{i\pi}{4}}},$$
$$\Large{z=e^{\frac{i3\pi}{4}}},\text{ and}$$
$$\Large{z=-e^{\frac{i3\pi}{4}}}$$
I am having trouble with the residues for each one. Are the answers just the poles but all divided by $4$? can someone help? Thanks!
 A: To find the residues you can use that
$$
\mathrm{Res}\left(\frac{f(z)}{g(z)},z_k\right) = \frac{f(z_k)}{g'(z_k)}
$$
if $z_k$ is a simple pole. To check that it is a simple pole
you only need to verify that the derivative at $z_k$ is not zero. 
Since you have found four different roots to a forth order polynomial, these are all simple. Or $g'(z_k) = 4 z_k^3 \neq 0$. 
Proof: If we have a simple pole at $z_k$ then $g'(z_k)\neq 0$ and $f(z_k)=g(z_k)=0$ and so
$$
\mathrm{Res}\left(\frac{f(z)}{g(z)},z_k\right) 
= \lim_{z\to z_k} \left(z-z_k\right) \frac{f(z)}{g(z)}
= f(z_k)\bigg/\lim_{z\to z_k} \frac{g(z)-g(z_k)}{z - z_k}
= \frac{f(z_k)}{g'(z_k)}
$$
A: Like N3buchadnezzar just said the residues are given by 
$$\mathrm{Res}\left(\frac{f(z)}{g(z)},z_k\right) = \frac{f(z_k)}{g'(z_k)}$$
In your case the algebra involved in the calculation may lead to many errors if you consider the residues as you listed them. I suggest you to write the singularities of $\frac{1}{z^4+1}$ as
\begin{align*}z_1 &= \frac{1+i}{\sqrt{2}} & z_2&=\frac{-1+i}{\sqrt{2}}\\
z_3&=\frac{1-i}{\sqrt{2}} & z_4&=\frac{-1-i}{\sqrt{2}}
\end{align*}
You can find these singularities just by a simple geometric reasoning. 
Now i think it's easier to evaluate the residues. For intance you get
\begin{align*}
\mathrm{Res}\left(\frac{1}{z^4+1},z_1\right) &= \mathrm{Res}\left(\frac{1}{z^4+1},\frac{1+i}{\sqrt{2}}\right)\\
&=\frac{1}{4z^3}\mid_{\frac{1+i}{\sqrt{2}}}\\
&=\frac{1}{\frac{4(1+i)^3}{\sqrt{2}}}\\
&=\frac{1}{\frac{4(-2+2i)}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}}\\
&=\frac{1}{4\sqrt{2}(-1+i)}\cdot \frac{(-1-i)}{(-1-i)}\\
&=\frac{-1-i}{8\sqrt{2}}
\end{align*}
You can evaluate the other residues very similar.
