Expand the Laurent Series for $f(z) = \frac{1}{z(z-i)^2}$ for $0 < |z - i| < 1$

My attempt

As $|z - i| < 1$, this means, that

$$\sum_{n=1}^{\infty} (z - i)^n = \frac{1}{1 - (z - i)}$$

But more than that, I don't really know what to do.

The problem here is, that what we get is:

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

So it's leading to nowhere basically. At least this approach. If it were instead $|z(z-2i)| < 1$, then it could work.

Maybe the job is here to show that indeed $|z(z-2i)| < 1$ if we have, that $|z-i| < 1$?

I can't find anything on the internet which could help me.

Solution 1:

It is a little easier to work around the origin, so define $w=z-i$. Then we have $|w|<1$ and so

$$\frac{1}{z}=\frac{1}{w+i}=-\frac{i}{1-iw}=-i\sum_{n=0}^\infty (-iw)^n$$

So

$$\frac{1}{z(z-i)^2}=\frac{-i}{w^2}\sum_{n=0}^\infty (-iw)^n$$