Find laurent series for function $\frac{z + i}{z^2}$ for $|z - i| > 1$

Question

So, I have function $$f = \frac{z + i}{z^2}$$ Since $f \in H(\{|z - i| > 1\})$ I was hoping to find laurent expansion. It is pretty easy to do it in case of writing down expansion for $|z - i| < 1$ with center at $i$ because in this case it is just its' Taylor series, I think. But I think I don't really understand how to create this expansion in case of $|z - i| > 1$, I guess at least it is easy to say that its'regular part should be all zero, otherwise norm of $f(z)$ goes to infinity as norm of $z$ goes to infinity (but in fact it goes to zero). I know that there exists general formula for coefficients in laurent expansion, I should just try to calculate an integral by curve, but I don't think that I want to do that, because there certainly is a simpler one. I am sorry for such stupid question, but I really couldn't do it myself for quite some time and want to understand how it can be done.

Answer 1

If $|z-i|>1$, then\begin{align}\frac1z&=\frac1{i+(z-i)}\\&=-\frac i{1-i(z-i)}\\&=-i\sum_{n=-\infty}^{-1}\bigl(-i(z-i)\bigr)^n\\&=\sum_{n=-\infty}^{-1}(-i)^{n+1}(z-i)^n\end{align}and therefore\begin{align}\frac1{z^2}&=-\sum_{n=-\infty}^{-1}n(-i)^{n+1}(z-i)^{n-1}\\&=\sum_{n=-\infty}^{-2}(n+1)(-i)^n(z-i)^n.\end{align}So,\begin{align}\frac{z+i}{z^2}&=\bigl(2i+(z-i)\bigr)\sum_{n=-\infty}^{-2}(n+1)(-i)^n(z-i)^n\\&=-2\sum_{n=-\infty}^{-2}(n+1)(-i)^{n+1}(z-i)^n+\sum_{n=-\infty}^{-2}(n+1)(-i)^n(z-i)^{n+1}\\&=-2\sum_{n=-\infty}^{-2}(n+1)(-i)^{n+1}(z-i)^n+\sum_{n=-\infty}^{-1}n(-i)^{n-1}(z-i)^n\\&=(z-i)^{-1}+\sum_{n=-\infty}^{-2}\bigl(-2(n+1)(-i)^{n+1}+n(-i)^{n-1}\bigr)(z-i)^n\\&=(z-i)^{-1}+\sum_{n=-\infty}^{-2}(3n+2)(-i)^{n-1}(z-i)^n.\end{align}

Answer 2

You can do this by partial fractions. Say $u=z-i$. Then $$f=\frac u{(u+1)^2}=\frac1{u+1}-\frac1{(u+1)^2}.$$Now if $|u|>1$ then $$\frac1{u+1}=\frac{\frac1u}{1+\frac1u}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{u^{n}}.$$

Differentiating that gives the Laurent series for $1/(u+1)^2$ and there you are.