Laurent series at z = 0

Question

I want to determine the Laurent series around z = 0 (so a Maclaurin series I think) of the following function

$f(z) = 4/(z^2+2z-3)$

which converges in $z = 1 + i$

I can rewrite the function as $f(z) = 4/(z(z+2)-3)$ but I can't do partial decomposition here...

And what should I do with the fact that it converges in $z = 1 + i$ ?

Answer 1

$$\frac4{(z+3)(z-1)}=-\frac1{3+z}-\frac1{1-z}=-\frac13\frac1{1+\frac z3}-(1+z+z^2+\ldots)=$$

$$-\frac13\left(1-\frac z3+\frac{z^3}9-\ldots\right)-(1+z+z^2+\ldots)$$

Do some order above now and justify steps.

Edit: Just was commented that you want convergence at $\;z=1+i\;$ , so let us leave $\;\frac1{1-z}\;$ in peace, and then pay attention to the fact that

$$\left|\frac{1+i}3\right|=\left|\frac{\sqrt2}3\right|<1$$

so that

$$-\frac1{3+z}=-\frac12\left(1-\frac z3+\ldots\right)$$

is valid for $\;z=1+i\;$ , and thus

$$\left.\frac4{z^2+2z-3}\right|_{z=1+i}=-\frac13\left(1-\frac{1+i}3+\frac{(1+i)^2}9-\ldots\right)+\overbrace{\frac1{1+i-1}}^{=-i}$$