LogarithmicFct

$\displaystyle 2^{i}$	$\displaystyle =$	$\displaystyle e^{i\log{(2)}} = e^{i [\log_{e}{\vert 2\vert} + i(0 + 2n\pi)]}$
	$\displaystyle =$	$\displaystyle e^{i [\log_{e}{\vert 2\vert}}e^{-2n\pi}$
	$\displaystyle =$	$\displaystyle e^{-2n\pi}[\cos{(\log_{e}{2})}+i \sin{(\log_{e}{2})}]$