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Given any complex number $z$, I am interested in investigating the convergence properties of $z$ to some power $n$. We have three cases:

i) If $|z| <1$, then we can write $z^n=\mathrm{e}^{nx}\mathrm{e}^{iny}$. Letting $n \to \infty$ we get that $\mathrm{e}^{nx}\to 0$, and since $\mathrm{e}^{iny}$ always has length one, $z^n \to 0$. Is there a more rigorous way of saying this?

ii) If $|z| > 1$, then for the same reasons $z^n \to \infty$.

iii) If $|z| =1$, then $z^n=\mathrm{e}^{iny}$ which for increasing $n$ just keeps rotating on the unit circle, and therefore the limit doesn't exist. $z=1$ is the only converging case. Right?

These arguments does not seem rigorous enough. Any ideas how to make this more rigorous?

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  • $\begingroup$ Poor $z=1$ has been forgotten. $\endgroup$ – André Nicolas Sep 8 '13 at 3:54
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For the rigorous version of (i): Let $\epsilon > 0$. Then $|z^n - 0| = |z^n| = e^{nx} < \epsilon$ for all large enough $n$, since $x < 0$.

For version (ii), it's similar.

For (iii), suppose that $z = e^{it}$ with $t \ne 0$ (that is, $z \ne 1$). Then $$|z^{n + 1} - z^n| = |e^{int + it} - e^{int}| = |e^{it} - 1|$$ does not converge to $0$, and so the limit does not exist.

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Hint: For all of then, just use $|z^n| = |z|^n$.

So if $|z|< 1$, then $\{|z|^n\}$ is a decreasing sequence of real numbers bounded below by $0$. Hence it must converge (check that the limit is zero).

If $|z| > 1$, then $w = 1/z$ satisfies $(i)$

If $|z| = 1$, you will have to use the irrationality of $\pi$

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  • $\begingroup$ I don't see how the irrationality of $\pi$ is relevant here. We can use that to show that $\{z^n\}$ is dense in $\{|z| = 1\}$ for $z \ne e^{ir}$ with $r$ rational using that fact, but the claim is much weaker. $\endgroup$ – user61527 Sep 8 '13 at 4:04

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