Test convergence of the series and find its sum: $\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$

Question

Test convergence of the series and find its sum: $$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$$

My try:

This is simplified to $$\sum_{n=0}^{\infty}\left(\frac{-i-8}{13}\right)^n$$

Then how to proceed?

Answer 1

Hint. One knows that, for $z \in \mathbb{C}$, $$ |z|<1 \implies\sum_{n=0}^{\infty}z^n \quad \text{is convergent} $$ and in this case $$ \sum_{n=0}^{\infty}z^n=\frac1{1-z}. $$ What is $\left|\dfrac{-1-2i}{2+3i} \right|$ ?

Answer 2

$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n $

Since $\frac{-1-2i}{2+3i} =\frac{-1-2i}{2+3i}\frac{2-3i}{2-3i} =\frac{-8-i}{13} $ and $|\frac{-8-i}{13}|^2 =\frac{65}{169} \lt 1 $, the sum converges.

Applying the standard formula, the sum is

$\begin{array}\\ \dfrac{1}{1-\frac{-8-i}{13}} &=\dfrac{13}{13+(8+i)}\\ &=\dfrac{13}{21+i}\\ &=\dfrac{13}{21+i}\dfrac{21-i}{21-i}\\ &=\dfrac{13(21-i)}{442}\\ &=\dfrac{21-i}{34}\\ \end{array} $