How can we prove that $z^{n+1} \to 0_{\mathbb{C}}$ as $n\to \infty$?

Question

In the book of Function of One Complex Variable by Conway, at page 31, it is given that

However, normally, if $z$ was a real number, we could argue that $z^{n+1}$ goes to zero as $n \to \infty$ if $|z| < 1$, but in the case of complex numbers, if $|z = a + bi| = a^2 + b^2 < 1$, then, for example, $$z^2 = a^2 - b^2 + (2ab)i,$$ and we can see that it might be the case that $2ab \geq b$, so if we see z = (a,b), then $z^2 = (a^2 - b^2, 2ab)$, and the statement $z^n \to 0_{\mathbb{C}}$ means both of the components of $z^n$ goes to $0_{\mathbb{R}}.

However, even when we square the number $z$, the components does not need to decrease, so how can we be sure that if $|z| < 1$, then $z^n \to 0_{\mathbb{C}}$ as $n\to \infty$ ?

tl:dr;

How can we prove that $z^{n+1} \to 0_{\mathbb{C}}$ as $n\to \infty$ ?

Answer 1

$z^{n+1} \to 0 \iff \lim |z^{n+1} - 0| = 0 \iff |z|^{n+1} \to 0.$

First iff: let $z^n = x_n +\mathrm i y_n$, then by $$ \max(|x_n|, |y_n|) \leqslant |z^n| \leqslant |x_n| + |y_n|, $$ we conclude the first iff.

Example: $z^2 = (a^2 - b^2) + \mathrm i 2ab$, then $a^2 -b^2 \leqslant a^2 + b^2$, $2ab \leqslant 2|ab| \leqslant a^2+b^2$.

Answer 2

Just write your complex numbers in polar form — are you familiar with the polar representation of complex numbers?

Write $z=re^{i\theta}=r(\cos\theta+i\sin\theta)$ with $r=|z|$. Then $z^{n}=r^{n}e^{in\theta}$; $|e^{in\theta}| = 1$ for all real $n,\theta$ thus

$$|z^{n}|=|r^{n}|\to0$$