# Radius of Convergence for a Complex Function

I'm really rusty on my series convergence, and I guess I was more asking for a bit of clarification on a question I'm working on.

I've been asked to find the power series expansion for $\frac{1}{3 - z}$ about the point $4i$, and to then give the radius of convergence.

After a bit of tedium, I got the power series expansion as follows;

$$\frac{1}{3 - z} = \frac{1}{3 - 4i - (z - 4i)}$$ $$= \frac{1}{3-4i} \frac{1}{1 - \frac{z-4i}{3-4i}}$$ $$= \sum_{n = 0}^{\infty} \frac{1}{3-4i} (\frac{z-4i}{3-4i})^n$$ $$= \sum_{n = 0}^{\infty} \frac{1}{(3-4i)^{n+1}} (z-4i)^n$$

Now, I know this will hold true for $\frac{|z - 4i|}{5} < 1$, which is equivalent to $|z-4i| < 5$, so is it correct to say that the radius of convergence for this function is 5 (ie; the ball of radius 5, centred at 4i)??

• Yes...and the work looks just fine. Nice. +1 May 31, 2014 at 12:35

The only pole of the function being at $3$ and the distance of $3$ and $4i$ being $5$ you can confirm the radius of convergence.