Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3 
I am trying to find the Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ with centre $1$ that converges at $3$.

I was thinking that if we want the series to converge at $3$, we are interested in the region $\sqrt{2}<|z-1|<\sqrt{5}$. Hence, 
\begin{align}
f(z)&=-\frac{1}{2i-1-(z-1)}+\frac{1}{-i-1-(z-1)} \\
&=-\frac{1}{2i-1}\left(\frac{1}{1-\frac{z-1}{2i-1}}\right)-\frac{1}{z-1}\left(\frac{1}{1-\frac{-1-i}{z-1}}\right) \\
&=-\frac{1}{2i-1}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^n-\frac{1}{z-1}\sum_{n=0}^{\infty}\left(\frac{-1-i}{z-1}\right)^n \\
&=-\sum_{n=0}^{\infty}\left(\frac{(z-1)^n}{(2i-1)^{n+1}}+\frac{(-1-i)^n}{(z-1)^{n+1}}\right)
\end{align}
Is this solution correct? This question is different to others I have done in the past.

$$
\begin{align}
f(z)
&=\frac1{z-2i}-\frac1{z+i}\\
&=\frac1{w+1-2i}-\frac1{w+1+i}\\
&=\frac1{1-2i}\sum_{k=0}^\infty(-1)^k\left(\frac w{1-2i}\right)^k
-\frac1w\sum_{k=0}^\infty(-1)^k\left(\frac{1+i}w\right)^k\\
&=-\sum_{k=0}^\infty\frac{w^k}{(-1+2i)^{k+1}}-\sum_{k=0}^\infty\frac{(-1-i)^k}{w^{k+1}}\\
&=-\sum_{k=0}^\infty\frac{(z-1)^k}{(-1+2i)^{k+1}}-\sum_{k=1}^\infty\frac{(-1-i)^{k-1}}{(z-1)^k}\\
\end{align}
$$
Your series looks correct and converges for $\sqrt2\lt|z-1|\lt\sqrt5$.
