In the answers given to this question, the following relation is often used:

$$\left| z^x \right| = \left| z \right|^x$$

with $z \in \mathbb{C}$, $z = \alpha + i \beta$.

  • How to prove it?
  • Can $x$ be a complex number itself, or should $x$ only be a rational number?
  • 2
    $\begingroup$ $i^i \neq 1^i$ so $x$ has to be real $\endgroup$ – Archis Welankar Feb 1 '16 at 8:48
  • $\begingroup$ @ArchisWelankar please, read the comment I wrote in the corindo answer. $\endgroup$ – BowPark Feb 1 '16 at 9:17
  • $\begingroup$ Someting seems to be wrong $\endgroup$ – Archis Welankar Feb 1 '16 at 9:19
  • $\begingroup$ @ArchisWelankar tried to fix it. $\endgroup$ – BowPark Feb 1 '16 at 9:22

You can write $z$ in the form $z = re^{i\theta}$ with $r \in \mathbb{R}_+, \theta \in \mathbb{R}$.

$$ \forall x \in\mathbb{R}, \,\, |z^x| = r^x = |z|^x.$$

As noticed @Archis Welankar, when $x$ is complex the relation is not always true.

  • $\begingroup$ Thank you. But for $x \in \mathbb{C}$ I can't see it. If $x = u + iv$, $$(z)^x = (z)^{u + iv} = (r e^{i \theta})^{u + iv} = r^u r^{iv} e^{i \theta u} e^{i \theta iv} = r^u r^{iv} e^{i \theta u} e^{- \theta v}$$ and...? $\endgroup$ – BowPark Feb 1 '16 at 9:21

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