<s>[INST] Why does $\sqrt[4]{-2}=\frac{1+i}{\sqrt[4]{2}}$? When I type $\sqrt[4]{-2}$ into WolframAlpha, it lists $\frac{1+i}{\sqrt[4]{2}}$ as an alternate form. How did it get that? I was trying to solve for $x^{4}+2=0$, and so I guess I could just say $\sqrt[4]{-2}$ is a solution, but I want it to look like $a+bi$ for $a,b\in\mathbb{R}$. 
I remember that the 4th roots of unity are $e^{\frac{2\pi k}{4}i}$ for $k=1,2,3,4$, but we want to get $-2$, not $1$.
 [/INST] $$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{(1+i)^4}{2}=\frac{1}{2}\left(\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)\right)^4=\frac{4}{2}\left(\frac{1+i}{\sqrt{2}}\right)^4=2\left(\exp\left(\frac{i\pi}{4}\right)\right)^4=$$
$$=2\exp(i\pi)=2*(-1)=-2$$
I've used these:


*

*$$\exp\left(\frac{i\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)=\frac{1+i}{\sqrt{2}}$$

*$$\exp(i\pi)=\cos(\pi)+i\sin(\pi)=-1$$


As @DonAntonio pointed it out in the comments, it's much easier this way:
$$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{1}{2}(1+i)^4=\frac{1}{2}((1+i)^2)^2=\frac{1}{2}(1^2+2i+i^2)^2=\frac{(2i)^2}{2}=-2$$
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