How to find if two series are analytic continuations of each other Show that the series $a) \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$ and $b) \sum_{n=0}^{\infty} \frac{(z-i)^n}{(2-i)^{n+1}}$ are analytic continuations of each other.
I have performed the ratio test for both series and found that $a)$ converges for $|z|<1$ and $b)$ for $|\frac{z-i}{2-i}|<1$. I have a hint that says $b)$ also converges for $|z-i|< \sqrt{5}$.
I am not sure where the second convergence of $b)$, ($|z-i|< \sqrt{5}$), comes from.
I am also not really sure what to do next. My notes are very vague. Wouldn't the convergence of $|\frac{z-i}{2-i}|<1$ not be an analytic continuation since $1 \ngtr 1$. But the second convergence would.
I don't know if I am saying that right,  but that is what I am getting from my notes.

For $|z|<2$, the series $\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$ represents the function $f(z)= \frac1{2-z}$. Then, note that we can write for $|z-i|<|2-i|=\sqrt 5$
$$\begin{align}
f(z)&=\frac{1}{2-z}\\\\
&=\frac{1}{(2-i)-(z-i)}\\\\
&=\frac1{2-i}\left(\frac1{1-\frac{z-i}{2-i}}\right)\\\\
&=\sum_{n=0}^\infty \frac{(z-i)^n}{(2-i)^{n+1}}
\end{align}$$
Hence, we see that $f(z)=\sum_{n=0}^\infty \frac{(z-i)^n}{(2-i)^{n+1}}$ for $|z-i|<\sqrt 5$ is indeed the analytic continuation of $f(z)=\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$ for $|z|<2$.
