So I want to calculate the Laurent series of this function

$$ f: \mathbb{C} \to \mathbb{C}, \quad f(z) = \frac{1}{z^{2}+1}.$$

The Laurent series has to be in this form:

$$\sum_{n=- \infty }^{ \infty } a_{n} (z-i)^n$$

for a circular disc $$ 0<| z-i|<p,$$ where $p$ has to be found.

With partial fraction expansion I am getting $$ f(z) =\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right).$$

For the first summand, $$\frac{1}{z+i} = \frac{1}{2i} \frac{1}{1+\frac{z-i}{2i}} = \frac{1}{2i} \sum_{n= 0 }^{ \infty }\left(\frac{-(z-i)}{2i}\right)^n = \frac{1}{2i} \sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n $$ for $$\left|\frac{-(z-i)}{2i}\right| < 1 \Longrightarrow \left| z-i \right| < 2.$$ Now I don't know how to continue with $$\frac{1}{z-i} .$$

  • $\begingroup$ The roots of the denominator are $\pm i$. Therefore, the partial fractions should have $z\pm i$ are denominators. $\endgroup$ – user530511 Feb 12 '18 at 13:21
  • $\begingroup$ Partial fraction decomp should be $$f(z)=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)=\frac{i}{2}\left(\frac{1}{z+i}-\frac{1}{z-i}\right)$$ $\endgroup$ – Dave Feb 12 '18 at 13:24
  • $\begingroup$ yes, sorry, i have written "1" for "i", but i meant "i" $\endgroup$ – Chopin Feb 12 '18 at 13:30

In my opinion it is easier without partial fraction decomposition: let $z=w+i$ then for $0<|w|<2$ $$f(z) = \frac{1}{z^{2}+1}=\frac{1}{w(w+2i)}=\frac{1}{2iw(1-iw/2)}=-\frac{i}{2w}\sum_{k=0}^{\infty}(iw/2)^k.$$ Hence the Laurent expansion of $f$ in $0<|z-i|<2$ is $$f(z)=-\frac{i}{2(z-i)}+\sum_{k=0}^{\infty}\frac{i^{k}(z-i)^{k}}{2^{k+2}}$$


Note: Despite the question at the end regarding how to continue note that already all calculations were done in order to solve the problem.

The function \begin{align*} f: \mathbb{C} \to \mathbb{C}, \quad f(z) &= \frac{1}{z^{2}+1}\\ &=\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right)\tag{1} \end{align*} is to expand in a Laurent series at $z=i$.

We observe in (1) that \begin{align*} -\frac{i}{2}\cdot\frac{1}{z-i}\tag{2} \end{align*} is the principal part of the Laurent series of $f$. On the other hand we know that \begin{align*} \frac{i}{2}\cdot\frac{1}{z+i}=\frac{1}{4} \sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n\tag{3} \end{align*} is the power series representation of $f$ at $z=i$ with region of convergence $|z-i|<2$.

We conclude from (2) and (3) the Laurent series expansion of $f$ around $z=i$ is \begin{align*} f(z)&=\frac{1}{z^{2}+1}\\ &\color{blue}{=-\frac{i}{2}\cdot\frac{1}{z-i}+\frac{1}{4}\sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n} \end{align*}


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