# If $\lvert z \rvert < 1$ then $z^n \to 0$ as $n \to \infty$

$$\renewcommand{\abs}{\lvert #1 \rvert}$$Let z be a complex number. I would like to show that if $$\abs{z} < 1$$, then $$z^n \to 0$$ as $$n\to\infty$$. Could anyone provide me with a proof of this? I have no trouble showing the corresponding statement if $$z$$ were a real number.

• Use $|z^n|=|z|^n$. – SarGe May 20 at 7:35

$$|z^{n}|=|z|^{n} \to 0$$ by the real case since $$|z|$$ is a real number in $$[0,1)$$. If $$\epsilon >0$$ then there exists $$n_0$$ such that $$|z|^{n} <\epsilon$$ for all $$n >n_0$$. Hence $$|z^{n}| <\epsilon$$ for all $$n >n_0$$. This is what it means to say that $$z^{n} \to 0$$.
• $\renewcommand{\abs}{\lvert #1 \rvert}$Yes, I got this far as well. But how does it follow that $z^n \to 0$? – Simon SMN May 20 at 7:32
• @SimonSMN If the norm of a sequence of complex number tends to $0$, the limit of the sequence is $0$. Think geometrically. – MathematicsStudent1122 May 20 at 7:34
Let $$z=re^{i\theta}$$. Then $$z^n=r^ne^{i\theta n}$$. As $$|e^{i\theta n}|=1$$, and $$r\lt1$$, then $$z^n\to0$$ as $$n\to\infty$$.