# How to find the sum of a geometric series that involves complex number $i$?

Suppose that I have the following sum: $$\sum_{m=0}^{\infty}(e^{it}(1-p))^{m}$$, where $$i^2 = -1$$.

This is a geometric series, but involving the complex number $$i$$. Can I just apply the geometric series formula and conclude that the sum if $$\frac{1}{1-e^{it}(1-p)}$$?

Thank you,

• Yes, as long as $|1-p|<1$. – Angina Seng Oct 14 '20 at 19:58

To prove that if $$|z|<1$$ then $$\sum_{m\ge0}z^m=\tfrac{1}{1-z}$$, note that $$\tfrac{1}{1-z}-\sum_{m=0}^{n-1}z^m=\frac{z^n}{1-z}$$ has $$n\to\infty$$ limit $$0$$. (By contrast, if $$|z|\ge1$$ then this $$n\to\infty$$ behaviour of the error term is lost, so the series diverges.) Note nothing about this reasoning cares whether $$z$$ is real or complex.