I missed a lecture last class and I have absolutely no idea how to even start this. I understood how to do radius of convergence normally perfectly fine but my book isn't helping me at all with power series, and i've already tried using my calc textbook for help. I am going to office hours tomorrow to get help but I would like to get some of my HW done so I don't go to his office and say "Hey I know absolutely nothing, show me how to do this".
a) $\frac{1}{z-1}$ about $z = i$.
For this one, I am observing that there will be a singularity at $|z|$, but in the book they do stuff with just filling in bad points and i'm not sure where to go from here.
b) $\frac{z - i}{z^3 - z}$ about $z = 2i$.
$$ \begin{aligned} \frac{1}{z-1} &= \frac{1}{z-i-(1-i)} \\ &=-\frac{1}{1-i} \cdot \frac{1}{1-\frac{z-i}{1-i}} \\ &=-\frac{1}{1-i} \sum_{n=0}^{\infty} \frac{1}{(1-i)^n}(z-i)^{n} \end{aligned} $$ Now it is clear the the radius of convergence is $$ \begin{aligned} R &= \frac{1}{\limsup \sqrt[n]{\frac{1}{|1-i|^{n}}}} \\ &= |1-i| \\ &= \sqrt{2} \end{aligned} $$
Can you use this method to find the answer for part $(b)$?
Edit:
To justify the first equality, simply add and subtract $i$ in the denominator. Then, from the first line to the second, I factored out a $-(1-i)$ from the denominator. The next step is the use the fact that $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}$ for $|z|<1$. Instead of $z$, here we are dealing with $\frac{z-i}{1-i}$. Finally, to find the radius of convergence, use the fact that $\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_{n}|}$, where $a_{n} = \frac{1}{(1-i)^{n}}$.