where does $\frac{1}{1-z}$ about the point $5i$ converge. Hi: Th next question in John D'Angelo's text is exercise 4.8: where does the series for $\frac{1}{1-z}$ about the point $5i$ converge ?
I understand that the expansion is : $\sum_{n=0}^{\infty} (z - 5i)^{n}$. 
Now, for the series to converge, $|z-5i|$ has to be less than 1 because the series is geometric. So is that the answer ? that $|z-5i|$ < 1$. This exercise is after another exercise which was much harder ( required abel's convergence for complex series test ) so I'm thinking that maybe I'm not correct. Thanks.
 A: You want to write
$${1\over 1-z}={1\over 1-5i-(z-5i)}$$
Then this is just
$${1\over 1-5i}\left({1\over 1-{z-5i\over 1-5i}}\right)$$
Which, by geometric series, instantly gives
$${1\over 1-5i}\sum_n {(z-5i)^n\over (1-5i)^n}=\sum_n{(z-5i)^n\over (1-5i)^{n+1}}$$
and moreover we know geometric series have convergence if and only if the common ratio, $r$, has $|r|<1$ i.e. for
$$\left|{z-5i\over 1-5i}\right|<1\iff |z-5i|<|1-5i|=\sqrt{26}$$
A: As was noted by Adam Hughes, the series for $\frac1{1-z}$ about the point $5i$ is
$$
\begin{align}
\frac1{1-z}
&=\frac1{(1-5i)-(z-5i)}\\
&=\frac1{1-5i}\frac1{1-\color{#C00000}{\frac{z-5i}{1-5i}}}\\
&=\frac1{1-5i}\sum_{k=0}^\infty\left(\color{#C00000}{\frac{z-5i}{1-5i}}\right)^k\\
&=\sum_{k=0}^\infty\frac{(z-5i)^k}{(1-5i)^{k+1}}
\end{align}
$$
which converges by the ratio test for $|z-5i|\lt|1-5i|=\sqrt{26}$.
Another indicator of the radius of convergence of the Taylor series is the distance from the center of the expansion to the nearest singularity. Since the only singularity is at $z=1$, the radius of convergence is
$$
|1-5i|=\sqrt{26}
$$
