Given $f(z) =\frac{z^2-1}{z^2+1}$, I need to find it's Laurent series expansion at open disk $\sqrt{2} < |z-1| < +\infty$
So at first I've found that, at $z=\pm i$ function is not defined.Yet I do not quite understand key steps here, so please correct me if I'm mistaken.
$$f(z) = 1 - \frac{2}{z^2+1}$$ I need to simplify this expression so I can use known expansion formulas $$L(z) = \frac{2}{z^2+1} = i [\frac{1}{z+i} - \frac{1}{z-i}] = i [\frac{1}{(z-1)+i+1} - \frac{1}{(z-1)-(i-1)}]$$
Here I have that $ |\frac{i+1}{z-1}| = |\frac{i-1}{z-1}| < 1$
So I can use known geometric series formula $$L(z) = \frac{i}{z-1} [\frac{1}{\frac{i+1}{z-1}+1} - \frac{1}{1-\frac{i-1}{z-1}}] = \sum_{n=0}^{\infty} \frac{i (-1)^n (i+1)^n}{(z-1)^{n+1}} - \sum_{n=0}^{\infty} \frac{i (i-1)^n}{(z-1)^{n+1}} = \sum_{n=0}^{\infty} \frac{i [(-1)^n (i+1)^n - (i-1)^n]}{(z-1)^{n+1}}$$
I hope I've not done any mistake till now. My major problem is, that final answer is unreachable, which is $1 + \sum_{n=1}^{\infty}(-1)^n \frac{2^{\frac{n+2}{2}} \sin(\frac{\pi n}{4})}{(z-1)^{n+1}}$. Any hints are much appreciated, thank you
You just have to write : $1+i = \sqrt{2} e^{i\pi/4}$ (and same style for the other one), and simplify your calculations.