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$\renewcommand{\abs}[1]{\lvert #1 \rvert}$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.

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  • $\begingroup$ Use $|z^n|=|z|^n$. $\endgroup$ – SarGe May 20 at 7:35
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$|z^{n}|=|z|^{n} \to 0$ 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 $z^{n} \to 0$.

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  • $\begingroup$ $\renewcommand{\abs}[1]{\lvert #1 \rvert}$Yes, I got this far as well. But how does it follow that $z^n \to 0$? $\endgroup$ – Simon SMN May 20 at 7:32
  • $\begingroup$ @SimonSMN If the norm of a sequence of complex number tends to $0$, the limit of the sequence is $0$. Think geometrically. $\endgroup$ – MathematicsStudent1122 May 20 at 7:34
  • $\begingroup$ Do you know the definition of limit for complex sequences? @SimonSMN $\endgroup$ – Kavi Rama Murthy May 20 at 7:35
  • $\begingroup$ Got it now, Kavi. Many thanks for your excellent explanation. I needed a reminder of what the definition of convergence for a complex sequence is. $\endgroup$ – Simon SMN May 20 at 7:38
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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 $n\to\infty$.

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Let z=modz e^i0 z^n=modz^n e^i0*n as e^i0n is some complex no. And mod z^n is 0 so 0 into any no. Is anyways zero

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