Exponant is a square root?

[Rictus]

Working with converting imaginary numbers and polar coordinates. Ran into a problem with a sqrt as the exponant. Can't seem to find anything on this and can't remeber doing this. I have to convert (sqrt(5) + 2i)^sqrt2 to the form re^i angle.
I just can't get started because the exponant as a root is throwing me.
Thanks,
Rick

 

[pka]

convert (sqrt(5) + 2i)^sqrt2 to the form re^i angle.

Recall that \(\displaystyle \L z^w = e^{w\left( {\log (z)} \right)} .\)


\(\displaystyle \L
\left( {\sqrt 5 + 2i} \right)^{\sqrt 2 } = e^{\sqrt 2 \left( {\log \left( {\sqrt 5 + 2i} \right)} \right)} = e^{\sqrt 2 \left( {\ln (3) + i\left( {Arg(\sqrt 5 + 2i) + 2\pi n)} \right)} \right)}\)