Laurent series for $\frac1{z^2+1}$

Question

I have this problem:

Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.

I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$, then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem. Do you know that series? Thank you.

Answer 1

Expanding geometrically, $$\frac{2}{z+2} = \frac{1}{ 1 + \frac z 2} = \sum_{k=0}^\infty (-1)^k \left( \frac z 2\right)^k $$ Similarly, $$\frac{2z-4}{z^2 + 1} = \frac{2z-4}{z^2}\cdot \frac 1 {1 + \frac{1}{z^2}} = \frac{2z-4}{z^2} \sum_{k=0}^{\infty}(-1)^k \left( \frac 1 {z^2}\right)^k$$

Answer 2

The answer by Santosh is not correct.

The partial fraction decomposition of $f(z)$ is $$ f(z) = \frac{2}{z+2} - \frac{1-2i}{z+i} - \frac{1+2i}{z-i} $$ Since we are looking for Laurent series in the annulus, $1<\lvert z\rvert < 2$, we have that $\frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$. \begin{align} \frac{2}{z+2} - \frac{1-2i}{z+i} - \frac{1+2i}{z-i} &= \frac{1}{1+\frac{z}{2}} - \frac{1-2i}{z}\frac{1}{1+\frac{i}{z}} - \frac{1+2i}{z}\frac{1}{1-\frac{i}{z}}\\ &=\sum_{n=0}^{\infty}(-1)^n\biggl[\Bigl(\frac{z}{2}\Bigr)^n-\frac{1-2i}{z^{n+1}}(i)^n\biggr]-(1+2i)\sum_{n=0}^{\infty}\frac{(i)^n}{z^{n+1}} \end{align}

Answer 3

Realize the last series you ask about as a geometric series... $$ \frac{1}{z^2+1} = \frac{1}{z^2}\;\frac{1}{1-(-1/z^2)} = \frac{1}{z^2} - \frac{1}{z^4} +\frac{1}{z^6} - \frac{1}{z^8}+\dots $$