Expand $\frac{1}{z-i}$ to power series around $z_0=1$ and find the radius of convergence
$$\frac{1}{z-i}=\frac{1}{(z-1)+1-i}=\frac{1}{1-i}\cdot\frac{1}{1+\frac{z-1}{1-i}}=\frac{1}{1-i}\cdot\frac{1}{1-(-\frac{z-1}{1-i})}=\frac{1}{1-i}\sum_{n=0}^\infty (-\frac{z-1}{1-i})^n=\frac{1}{1-i}\sum_{n=0}^\infty \frac{(-1)^n}{(1-i)^n}(z-1)^n=\sum_{n=0}^\infty \frac{(-1)^n}{(1-i)^{n+1}}(z-1)^n$$
If the above is correct how should I find the radius of convergence? I need to bring it to $\sum a_nz^n$ and use $\lim_{n\to\infty}|\frac{a_n}{a_{n+1}}|$?
$\endgroup$ 12 Answers
$\begingroup$An approach different from using the ratio test. Observe that when you used:
$$ \frac{1}{1-a} = \sum_{n=0}^\infty a^n $$
you assumed that $|a| < 1$. In your case:
$$ |\frac{z-1}{1-i}|<1 $$
a region in $\mathbb{C}$ that has radius $\sqrt{2}$.
$\endgroup$ 2 $\begingroup$What you did is correct. Now\begin{align*}\lim_{n\in\mathbb N}\left|\frac{\frac{(-1)^{n+1}}{(1-i)^{n+2}}}{\frac{(-1)^n}{(1-i)^{n+1}}}\right|&=\lim_{n\in\mathbb N}\frac{\sqrt2^{n+1}}{\sqrt{2}^{n+2}}\\&=\frac1{\sqrt2}.\end{align*}Therefore, the radius of convergence is $\sqrt2$.
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