$\left(\frac{-\sqrt{2}}{2}-\frac{1}{\sqrt{-2}}\right)^{2016}$

Question

This is a question I'm getting conflicting answers on.

Method 1: Rationalize denominator of $−1/\sqrt{−2}$ to get $+(\sqrt{2}/2)i$, then square the whole thing, resulting in $-i$, then $(-i)^{1013}$, getting $-i$ as a final answer.

Method 2: Rationalize like before to get in form $a + bi$, then convert to polar form, $r = 1$, so $(1(\cos\theta + i\sin\theta))^{2016}$, then $\theta$ is either $3\pi/4$ or $\pi/4$. Use De Moivre's to get $1(\cos(2016\theta) + i\sin(2016\theta))$. This results in $1 + 0i$.

Any help would be greatly appreciated.

Answer 1

When you squared the whole thing in method 1, you halved $2016$ to give $1013$, which is where the mistake is. It should be $1008$. This then does indeed give $1$, in agreement with method 2.

Answer 2

$$\left(\frac{-\sqrt{2}}{2}-\frac{1}{\sqrt{-2}}\right)^{2016}=\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)^{2016}$$ then use the fact $$\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)^2=-i$$