# Clarification of Step in Needham's $Visual$ $Complex$ $Analysis$

I'm working my way through Needham's Visual Complex Analysis. In $\S 2.3$ he talks about complex power series. He wants to generalize the geometric series; $$\sum_{j=0}^{\infty} x^j = 1 + x + x^2 + \dots$$ by expanding $1/(a-x)$ about $k$:

$$\frac{1}{a-x}=\frac{1}{a-(X+k)} = \frac{1}{a-k}\frac{1}{1-\frac{X}{a-k}}$$

where $X = (x-k)$ measures displacement of $x$ from the centre $k$. He then immediately claims

$$\frac{1}{a-x}=\sum_{j=0}^{\infty}\frac{X^j}{(a-k)^{j+1}} iff \hspace{3mm} \lvert X \lvert < \lvert a-k \lvert$$

I don't understand the last jump. Any clarification would be immensely appreciated! Thanks!

-
+1 for studying that book BTW. It's nice. –  WimC Nov 28 '12 at 20:12
I'm loving it. A real treat to read. –  user45793 Nov 29 '12 at 4:28
add comment

## 2 Answers

If

$$\left| \frac{X}{a - k} \right| < 1$$

then

$$\frac{1}{1 - \frac{X}{a - k}} = \sum_{j = 0}^{\infty} \left(\frac{X}{a - k} \right)^j.$$

This is just the geometric series.

-
lol righht. missed that. Thanks! –  user45793 Nov 27 '12 at 18:26
add comment

Well, you have that $\frac{1}{a-x}= \frac{1}{a-k}\frac{1}{[1-(\frac{X}{a-k})]}$. But this is exactly $\frac{1}{a-x}=\frac{1}{a-k}\sum_{j=0}^{\infty}(\frac{X}{(a-k)})^j$, using that $\sum_{j=0}^{\infty} x^j = \frac{1}{1-x}$ and $x:=\frac{X}{a-k}$.

The "iff" statement is obvious from $\sum_{j=0}^{\infty} x^j=\frac{1}{1-x}$ iff $|x|<1$. So, in your case, $|\frac{X}{(a-k)}|$<1, i.e. $\lvert X \lvert < \lvert a-k \lvert$

-
add comment