Finding the Laurent series when a residue at $z=0$ is given

I have $a_n=\frac{1}{n^4+1}$, where n is an integer. I want to determine the Laurent series for a function f(z) such that the residue for $\frac{f(z)}{z^n}$ in $z=0$ is $a_n \forall \ n$.

Based on this introduction, I start with that:

\begin{equation} Res_{z=0}\frac{f(z)}{z^n}=\frac{1}{n^4+1} \end{equation}

and:

\begin{equation} Res_{z=0} \frac{f(z)}{z^n}=\lim_{z\longrightarrow 0}(z-0)\frac{f(z)}{z^n}=\frac{1}{n^4+1} \end{equation}

I write it out as:

\begin{equation} Res_{z=0} \frac{f(z)}{z^n}=\lim_{z\longrightarrow 0}(z-0)\frac{f(z)}{(z-0)^n}=\lim_{z\longrightarrow 0}\frac{f(z)}{(z)^{n-1}}=\frac{1}{n^4+1} \end{equation}

Then I take that this equation (eqn. 1)

\begin{equation} \lim_{z\longrightarrow 0}\frac{f(z)}{(z)^{n-1}}=\frac{1}{n^4+1} \end{equation}

is the key to find the Laurent series (eqn.2 ):

\begin{equation} f(z)=...+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{(z-z_0)}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+...a_n(z-z_0)^n \end{equation}

But how can I solve this using eqn. 1 and 2?

UPDATE:

I use the formula for the Laurent Series:

\begin{equation} f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^n=\sum_{-\infty}^{\infty}\frac{1}{n^4+1}(z-z_0)^n \end{equation}

Since at residue $z=0$ we have $\frac{f(z)}{z}$, I have the following:

\begin{equation} \frac{f(z)}{z}=\sum_{-\infty}^{\infty}\frac{1}{n^4+1}\frac{(z-z_0)^n}{(z-z_0)}=\sum_{-\infty}^{\infty}\frac{1}{n^4+1}(z^{n-1}) \end{equation}

This would give the proposed form of the Laurent series with residue z=0 as given:

\begin{equation} \sum_{-\infty}^{\infty}\frac{(z^{n-1})}{n^4+1} \end{equation}

Inserting for $k=n-1$ we get

\begin{equation} \sum_{-\infty}^{\infty}\frac{(z^{k})}{(k+1)^4+1} \end{equation}

Would that be correct?

Solution 1:

If the Laurent series of $f(z)$ around $0$ is $ \sum\limits_{k=-\infty}^{\infty} b_kz^{k}$ then the Laurent series for $\frac {f(z)} {z^{n}}$ is $ \sum\limits_{k=-\infty}^{\infty} b_kz^{k-n}$ and the residue at $z=0$ is simply the coefficient of $\frac 1 z$. So $a_n=b_{n-1}$ Thus $b_n=a_{n+1}=\frac 1 {(n+1)^{4}+1}$ for all $n$.