Laurent Expansion Of $\frac{1}{z^2+1}$

Question

Expand $\frac{1}{z^2+1}$ around $0<|z-i|<2$

$$\frac{1}{z^2+1}=\frac{1}{2i(z-i)}-\frac{1}{2i(z+i)}=\frac{1}{2i}\cdot\frac{1}{(z-i)}-\frac{1}{2i}\cdot\frac{1}{(z+i)}$$

How can I expand it to Laurent series?

Answer 1

We have the complex function with respect to $z$ :

$$f(z) = \frac{1}{z^2+1}$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = \frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)}=\frac{1}{[(z-i)+2i](z-i)}$$ $$=$$ $$\frac{1}{(z-i)^2\big[1+\frac{2i}{z-i}\big]}$$

Recall now, that :

$$\frac{1}{1+w} = \sum_{n=0}^\infty (-1)^nw^n, \; \; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = \frac{1}{(z-i)^2}\cdot \frac{1}{1+\frac{2i}{z-i}}=\frac{1}{(z-i)^2}\sum_{n=0}^\infty(-1)^n\bigg(\frac{2i}{z-i}\bigg)^n$$ $$\implies$$ $$f(z) = \sum_{n=0}^\infty(-1)^n\frac{(2i)^n}{(z-i)^{n+2}}$$

for $\bigg| \frac{2i}{z-i} \bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.