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.

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
$$\left|{z-5i\over 1-5i}\right|<1\iff |z-5i|<|1-5i|=\sqrt{26}$$
