# Complex Analysis: Isolated Singularities, Poles, and Residues

I was given the following question.

Show that the isolated singularities of the function $f(z) = \frac{z}{z^4+4}$ are poles. Determine the order of each pole and find the corresponding residues.

I am familiar with the three types of isolated singularities we have: removable, essential, and poles. I am interested in determining the poles this function may have and then find the corresponding residue.

This was my approach to find the poles. By the fundamental theorem of algebra, we can assert that the denominator of the expression has four zeroes that make the function not analytic.

\begin{eqnarray} f(z) & = & \frac{z}{z^4+4}\\ & = & \frac{z}{(z^2+2i)(z^2-2i)}\\ & = & \frac{z}{(z-(1+i))(z-(-1-i))(z-(-1+i))(z-(1-i))} \end{eqnarray} This is where I want to conclude I have 4 simple poles at $z = 1+i, -1-i, 1-i, -1+i$. Is this valid?

If so, how would I continue to find the residue for each simple pole? I know there are three ways to find the residue, Res($f;z_o$) = $c_{-1}$.

If $f$ has a simple pole at $z_0$, then \begin{equation} c_{-1} = \lim_{z\rightarrow z_0}(z-z_0)f(z) = \frac{A(z_0)}{B'(z_0)} \end{equation} where $f(z) = \frac{A(z)}{B(z)}$; $A$, $B$ are analytic at $z_0$, $A(z_0) \neq 0$ and $B$ has a simple pole (zero) at $z_0$

My other choice in finding the residues is using Laurent Series but that route seems impractical in this situation.

Thus far, am I on the right track in finding the residues?

Thank you for your time and Thanks in advanced for any feedback.

Even if you haven't found all of the roots you can still argue if a given (or already found) root is a simple pole. The idea is computing $B'(z_0)$ at that point. Remember that a zero $z_0$ of a polynomial $P$ is a zero of order $k$ iff $P(z_0)=...=P^{(k-1)}(z_0)=0$ and $P^{(k)}(z_0)\neq0$. So, in the process of using the formula you quoted you already are checking that the pole is simple.