My attempt
We know, that $$\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$$ for $|z| < 1$
As we have $1 < |z| < \infty$, this means that we have such $z$ that $\left|\frac{1}{z}\right| < 1$, thus what we're dealing with is: $$\sum_{n=0}^{\infty} \left(\frac{1}{z}\right)^n = \frac{1}{1-\left(\frac{1}{z}\right)}$$
So we now have:
$$f(z) = \frac{1}{z(z-i)} = \frac{1}{z} \frac{1}{z - i} = \frac{1}{z} \frac{1}{z\left(1 - \frac{i}{z}\right)} = \frac{1}{z^2} \frac{1}{1 - \frac{i}{z}} = \frac{1}{z^2} \sum_{n = 0}^{\infty} \left(\frac{i}{z}\right)^n = \sum_{n=0}^{\infty} \frac{i^n}{z^{n + 2}}$$
Is this the solution? Are there any mistakes?
We haven't been given a proper introduction into how to find the Laurent Series of a function, all I know is from looking at examples but I'm not sure if I understood it properly how to do it. I couldn't find an example involving the imaginary unit anywhere, too.