Proof attempt:
$|z^n|=|z\cdot z \cdot \cdot \cdot z|=|z|\cdot|z|\cdot \cdot \cdot|z|=|z|^n$
So the property is true only for $n\in \mathbb{N}$?
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4Yes, this is correct. – Guy Jun 14 '17 at 18:49
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@Guy : No it isn't. See my answer. – MPW Jun 14 '17 at 18:57
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2That would only be correct if you first proved that $|a \cdot b|=|a| \cdot |b|$ and used that as the base case for a proof by induction. – dxiv Jun 14 '17 at 19:03
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It is actually true for all integers $n$, at least of $z\neq 0$. – Thomas Andrews Jun 14 '17 at 19:18
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@ThomasAndrews and $z^{-n}=\frac{1}{z^n}$ right? – gbox Jun 14 '17 at 19:30
2 Answers
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In general, for nonzero complex numbers $z^n = \exp(n \log(z))$, using any branch of the logarithm of $z$.
Thus $$ \left| z^n \right| = \exp(\text{Re}(n \log(z)))$$ while $$ |z|^n = \exp(n \log |z|) = \exp(n \text{Re}(\log(z)))$$
The two are equal if and only if $\text{Re}(n \log(z)) = n \text{Re}(\log(z)) + 2 m \pi i$ for some integer $m$. There are two cases where this occurs:
- $n$ is real.
- $\log(z)$ is real, i.e. $z$ is real and positive, with $\text{Im}(n) = - 2 m \pi/\log(z)$.
Robert Israel
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If we take any branch of the log of $z$ we can use the property of $log(z)^n=nlog(z)$? – gbox Jun 14 '17 at 19:17
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No, this is not correct.
You have assumed what you are trying to prove in the second equality.
You should use an inductive proof, after proving that $|z_1z_2|=|z_1||z_2|$.
MPW
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2You need to assume $|ab|=|a|\cdot |b|$, @MPW. $|z^2|=|z|^2$ is not enough to get this result. – Thomas Andrews Jun 14 '17 at 19:31
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