Aus Online Mathematik Brückenkurs 2

The calculation follows a fairly set pattern. We write the number \displaystyle 4\sqrt{3}-4i in polar form and then use de Moivre's formula.

This gives

\displaystyle 4\sqrt{3}-4i=8\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right)

and then we get, on using de Moivre's formula,

\displaystyle \begin{align} & \left( 4\sqrt{3}-4i \right)^{22}=8^{22}\left( \cos \left( 22\centerdot \left( -\frac{\pi }{6} \right) \right)+i\sin \left( 22\centerdot \left( -\frac{\pi }{6} \right) \right) \right) \\ & =\left( 2^{3} \right)^{22}\left( \cos \left( -\frac{11\pi }{3} \right)+i\sin \left( -\frac{11\pi }{3} \right) \right) \\ & =2^{3\centerdot 22}\left( \cos \left( -\frac{12\pi -\pi }{3} \right)+i\sin \left( -\frac{12\pi -\pi }{3} \right) \right) \\ & =2^{66}\left( \cos \left( -4\pi +\frac{\pi }{3} \right)+i\sin \left( -4\pi +\frac{\pi }{3} \right) \right) \\ & =2^{66}\left( \cos \left( \frac{\pi }{3} \right)+i\sin \left( \frac{\pi }{3} \right) \right) \\ & =2^{66}\left( \frac{1}{2}+i\frac{\sqrt{3}}{2} \right)=2^{65}\left( 1+i\sqrt{3} \right) \\ \end{align}