# Show that for any complex number $z$, $|r^z| = r^x$, where $r$ in the real numbers, $r>0$.

The question asks for the magnitude of $r^z$, and I take it that the $x$ will be from the fact that $z$ can be written as $z=x+iy$.

Been stuck on this for a while now, it's only worth 2 marks, I think I'm missing something glaringly obvious. Thanks.

• Hint: $r^z = e^{z\log r}$. – Antonio Vargas Jan 11 '14 at 17:48
• @AntonioVargas: that is choosing the main branch of the complex logarithm. – DonAntonio Jan 11 '14 at 17:49
• @DonAntonio, I didn't write $\mathrm{Log}$, it's $\log$. – Antonio Vargas Jan 11 '14 at 17:53
• It never mind what you wrote, @AntonioVargas: it must be the complex log as we're dealing with a complex function. That the logarithm's argument is real makes no difference. – DonAntonio Jan 11 '14 at 17:55
• Thanks Antonio, I've got it now! – Joe John Jan 11 '14 at 17:55

Choosing the main branch of the complex logarithm (taking out the non-positive real axis, and the one for which the argument of positive real numbers is zero), and putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , we get
$$r^z=e^{z\,\text{Log}\,r}=e^{(x+iy)\log r}=e^{x\log r}e^{iy\log r}\implies$$
$$|r^z|=\left|e^{x\log r}e^{iy\log r}\right|=e^{x\log r}=r^x$$
since we know $\;|e^{i\phi}|=1\;\;\forall\,\phi\in\Bbb R\;$