Complex conjugate of $x^{-i\Omega}$, for $x,\Omega>0$

Question

If $x>0$ and $\Omega>0$, what is $\left( x^{-i\Omega} \right)^{\ast}$?

I am 99% sure that it's $x^{i\Omega}$, but I can't see why this is the case.

Answer 1

Yes, since $x=e^{\ln(x)}$ and then $$(x^{-i\Omega})^*=(e^{-i\Omega \ln(x)})^*=(\cos(-\Omega \ln(x))+i\sin(-\Omega\ln(x)))^*=\cos(-\Omega \ln(x))-i\sin(-\Omega \ln(x))=\cos(\Omega \ln(x))+i\sin(\Omega\ln(x))=e^{i\Omega \ln(x)}=x^{i\Omega}$$