Find $\left(\frac{1+i}{\sqrt{2}}\right)^{46}$.
Let's rewrite the complex number in polar form. That means we need to find the magnitude and the argument. The magnitude of $1+i$ is $\sqrt{1^2+1^2}=\sqrt{2}$. The argument of $1+i$ is just $\tan^{-1}\left(\frac{1}{1}\right)=\tan^{-1}(1)=\frac{\pi}{4}$. So, the polar form of $1+i$ is $\left(\sqrt{2},\frac{\pi}{4}\right)$. The magnitude of $\frac{1+i}{\sqrt{2}}$ is just $\frac{\sqrt{2}}{\sqrt{2}}=1$. The argument of $\frac{1+i}{\sqrt{2}}$ is $\frac{\pi}{4}$. Now we can write $\frac{1+i}{\sqrt{2}}$ as $\left(1,\frac{\pi}{4}\right)$. Using DeMoivre's Theorem, we can find $\left(\frac{1+i}{\sqrt{2}}\right)^{46}$. Which is just $\left(1,46\cdot\frac{\pi}{4}\right)=\left(1,\frac{23\pi}{2}\right)$. That is equal to $-i$.

