I need help finding Laurent Series near a singularity at $z_0$ that converges for $0<|z-z_o|<R$ for $$f(z) = \frac{1}{z^2(z-i)}$$ and $z_0 = i$
The way I tried to solve it was to use the common Maclauren series: $$\tag{1} \frac{1}{z-a} = -\frac{1}{a} \sum_{k=0}^\infty \left(\frac{z}{a}\right)^n, \qquad |z| \leq a $$ and $$\tag{2} \frac{1}{z-a} = \frac{1}{z} \sum_{k=0}^\infty \left(\frac{a}{z}\right)^n, \qquad |z| \geq a.$$
I attempted: \begin{align*} f(z) &= \, \frac{1}{z^2} \frac{1}{i(\frac{z}{i}-1)} \\ &= \frac{-1}{z^2i} \frac{1}{1 - \frac{z}{i}} \\ \end{align*}
\begin{align*} \frac{1}{z^2(z-i)} &= \frac{-1}{z^2i} \sum_{n=0}^\infty \left(\frac{z}{i}\right)^n \\ &= -\sum_{n=0}^\infty \left(\frac{z^{n-2}}{i^{n+1}}\right) \end{align*}
Then using the Cauchy-Hadamard equation, I get the radius of convergence as $0<|z-i|<\infty$. I know this answer is wrong because it is not what is in the back of the book. What am I doing wrong here?
You're after a Laurent series centered at $i$, but you got one that's centered at $0$ instead.
Note that\begin{align}\frac1z&=\frac1{i+(z-i)}\\&=\frac{-i}{1-i(z-i)}\\&=-i\sum_{n=0}^\infty i^n(z-i)^n\text{ (if $|z-i|<1$)}\\&=-\sum_{n=0}^\infty i^{n+1}(z-i)^n.\end{align}Therefore\begin{align}\frac1{z^2}&=-\left(\frac1z\right)'\\&=\sum_{n=1}^\infty ni^{n+1}(z-i)^{n-1}\\&=\sum_{n=0}^\infty(n+1)i^{n+2}(z-i)^n.\end{align}It follows from this that\begin{align}\frac1{z^2(z-i)}&=\sum_{n=0}^\infty(n+1)i^{n+2}(z-i)^{n-1}\\&=\sum_{n=-1}^\infty(n+2)i^{n+3}(z-i)^n.\end{align}