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Example Express the following in the form of $a + bi$．

$\displaystyle 2^{i}$

Answer 1. We use $a^{z} = e^{z\log{a}}$．


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