Finding the Laurent series of
$$f(z) = \frac{-z^2-2z+3i}{z-2i}$$
on the annulus of infinite external radius $|z-i| > 1$
This function looks fairly messy. We clearly have a simple pole at $z=2i$. But it's not obvious to me how to find the principal and analytic part given this particular numerator. Any advice on how to obtain this expansion? Thanks
UPDATE: I attempted to rewrite this using long division. When doing so, I obtain
$$(-z^2-2z+3i) \div (z-2i) = -z-2-2i-\frac{i-4}{z-2i}$$
In other words, I obtained a remainder of $i-4$.
Not sure if this is helpful.
Note that, if $|z-i|>1$,\begin{align}\frac1{z-2i}&=\frac1{-i+(z-i)}\\&=\frac i{1+i(z-i)}\\&=-i\sum_{n=-\infty}^{-1}\bigl(-i(z-i)\bigr)^n\text{ (because $\bigl|i(z-i)\bigr|>1$)}\\&=-i\sum_{n=-\infty}^{-1}(-i)^n(z-i)^n\\&=\sum_{n=-\infty}^{-1}(-i)^{n+1}(z-i)^n.\end{align}Can you take it from here?