Find the Taylor series for $f(z)=\frac{i}{(z-i)(z-2i)}$ about $z_0=0$ and the disk of convergence.
For the Taylor series I got
$-\frac{i}{2}+\frac{z(2+i)}{4}-\frac{z(2i+3)}{2!}…$, but I'm not super confident in it. Can someone confirm or deny if it's correct?
To find this I simply used the $๐(๐ง_0)+\frac{๐โฒ(๐ง_0)}{1!}(๐งโ๐ง0)+\frac{๐โณ(_0)}{2!}(๐งโ๐ง_0)^2+โฏ$ expansion. I double checked with a derivative calculator so I know my values are right, I just wanted to double check that this is the right way to do it. I always thought it was, but that's not how my complex analysis class does it and I don't understand their way.
Best Answer
Note that, if $z\in\Bbb C\setminus\{i,2i\}$,\begin{align}\frac i{(z-i)(z-2i)}&=\frac1{i-z}-\frac1{2i-z}\\&=-\frac i{1+iz}+\frac12\frac i{1+iz/2}.\end{align}Therefore, if $|z|<1$,\begin{align}\frac i{(z-i)(z-2i)}&=-i\sum_{n=0}^\infty(-iz)^n+\frac i2\sum_{n=0}^\infty\left(-\frac{iz}2\right)^n\\&=\sum_{n=0}^\infty\left((-i)^{n+1}-\left(-\frac i2\right)^{n+1}\right)z^n\\&=\sum_{n=0}^\infty(-i)^{n+1}\left(1-\frac1{2^{n+1}}\right)z^n.\end{align}So, this last power series is the Taylor series that you're after.