While studying I got this exercise:
Find the Laurent Series expansion valid for $0 < |z – i| < \sqrt2$ for the following function:$$f(z) = \frac{1}{(z-i)^8(z+1)}$$
So I have to get a series expansion around the point $i$.
I tried using $w = z – i$ to see if I could get somewhere but all that I got was $$f(w) = \frac{1}{w^8}*\frac{1}{w + i + 1}$$ which doesn't help me much because I can't turn the second fraction into a geometric series, I thought that maybe I was supposed to use the $|1 + i|$ especially because that evaluates to $\sqrt2$ but I don't really know if there is a valid way to do that.
Appreciate any help.
Edit: From AndreasBlass comment I got it to $$\frac{1}{i+1}\sum_{n=0}^\infty \frac{w^{n-8}}{(1+i)^n}$$ which I think would indeed be valid for the range they want, because as I said $|1 + i| = \sqrt2$
Best Answer
As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$\frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$\frac{1}{w^8(1+i)}\frac{1}{1 + \frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.