Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3

Question

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.

Answer 1

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