Aus Online Mathematik Brückenkurs 2

Because we are going to raise something to the power \displaystyle \text{12}, the base in the expression should be written in polar form. In turn, the base consists of a quotient which it is advantageous to calculate in polar form. Thus, it seems appropriate to write \displaystyle 1+i\sqrt{3} and \displaystyle \text{1}+i in polar form right from the beginning and to carry out all calculations in polar form.

We obtain

\displaystyle \begin{align} & 1+i\sqrt{3}=2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\ & \text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\ \end{align}

and

\displaystyle \begin{align} & \frac{1+i\sqrt{3}}{\text{1}+i}=\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}{\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)} \\ & =\frac{2}{\sqrt{2}}\left( \cos \left( \frac{\pi }{3}-\frac{\pi }{4} \right)+i\sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) \right) \\ & =\sqrt{2}\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right) \\ & \\ & \\ & \\ \end{align}

Finally, de Moivre's formula gives

\displaystyle \begin{align} & \left( \frac{1+i\sqrt{3}}{\text{1}+i} \right)^{12}=\left( \sqrt{2} \right)^{12}\left( \cos 12\centerdot \frac{\pi }{12}+i\sin 12\centerdot \frac{\pi }{12} \right) \\ & =2^{\frac{1}{2}\centerdot 12}\left( \cos \pi +i\sin \pi \right) \\ & =2^{6}\centerdot \left( -1+i\centerdot 0 \right) \\ & =-64 \\ \end{align}