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 $1$.
 A: We could try writing $-2$ in polar form. Explicitly
\begin{equation}
 - 2 = 2 e^{i \pi}.
\end{equation}
Now taking the fourth root,
\begin{equation*}
 (-2)^{1/4} = 2^{1/4}(e^{i \pi})^{1/4}.
\end{equation*}
Can you take it from there?
A: Note that $\frac{1+i}{\sqrt[4]2} = \frac1{\sqrt[4]{2}} + \frac1{\sqrt[4]{2}}i$, so while it isn't written on the form $a+bi$, it is really easy to transform it so that it is.
As for why $\sqrt[4]{-2}$ is given as that number, try raising $\frac{1+i}{\sqrt[4]2}$ to the fourth power, trudge through the arithmetic and you will get $-2$.
If you are wondering how to solve such a problem on your own, I would suggest doing it geometrically, in polar coordinates (although it's entirely possible to do without writing any powers of $e$). We want a number with an argument which is a quarter of the argument of $-2$. The argument of $-2$ is $\pi$, so $\frac\pi4$ will do. (Thinking back to the $a+bi$ form, this just means that $a = b$, and they're both positive.)
Next we need the length to be the fourth root of the length of $-2$. The length of $-2$ is $2$, so the length of our fourth root must be $\sqrt[4]{2}$. This means that $\sqrt{a^2 + b^2} = \sqrt[4]2$, or since $a = b$, $2a^2 = \sqrt2$.
Solving this we get $a = b = \frac1{\sqrt[4]2}$, and there is our solution.
Now, don't forget that any non-zero complex number has four fourth roots, so you really need to find the other three. The implicit assumption that $\sqrt[k]{\hphantom x}$ gives a single answer is a big reason why I (and many others) don't like to use the symbol when dealing with complex numbers. In fact, even defining $i$ as $i = \sqrt{-1}$ is wrong in my opinion. It should be $i^2 = -1$.
A: Although polar form is by far a better way to compute complex roots, you can try this for this problem:
A fourth root of $-2$ is a square root of $\pm\sqrt 2 i$.
Say that
$$(a+ib)^2=\pm\sqrt 2i$$
Then, taking real and imaginary parts respectively,
$$a^2-b^2=0$$
$$2ab=\pm\sqrt 2$$
Solve the system for $a$ and $b$ and you should find the Wolfram's solution (and three more). Note that this system is actually two systems, because of the $\pm$ sign.
A: $$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{(1+i)^4}{2}=\frac{1}{2}\left(\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)\right)^4=\frac{4}{2}\left(\frac{1+i}{\sqrt{2}}\right)^4=2\left(\exp\left(\frac{i\pi}{4}\right)\right)^4=$$
$$=2\exp(i\pi)=2*(-1)=-2$$
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:
$$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{1}{2}(1+i)^4=\frac{1}{2}((1+i)^2)^2=\frac{1}{2}(1^2+2i+i^2)^2=\frac{(2i)^2}{2}=-2$$
A: It's worth noting that this only gives the principal value – since the $n$th root of complex number has $n$ complex roots, then this is only one of the values of $(-2)^{1/4}$.
That being said, rearranging your equation gives $(-2)^{1/4} \cdot 2^{1/4} = 1+i$. When you take powers in polar form, you take the powers of the lengths and multiply the angles (from De Moivre's theorem); so when you take roots, you take the roots of the lengths and divide the angles.
Using this,  $-2$ has $r = 2$ and $\theta = \pi$, so the principal value of $(-2)^{1/4}$ has $r = -2$ and $\theta = \frac{\pi}{4}$. Similarly, $2^{1/4}$ has $r=2^{1/4}$ and $\theta = 0$. Therefore, using polar multiplication $(r = 2^{1/4}, \theta = \frac{\pi}{4}) \cdot (r = 2^{1/4}, \theta =0)$ has $r = 2^{1/2}, \theta = \frac{\pi}{4}$; which is $1+i$ in Cartesian form.
A: This equality is $75\%$ wrong. 
Any non-zero complex number has exactly four fourth roots, and $z$ is one of them then the others are given by $z\zeta_4^k$ where $\zeta_4=e^{2i\pi/4}$ and $k=0, 1, 2, 3$.
Therefore the left-hand side does not represent one number without ambiguity, but four. You can check that the right-hand side is one of them (*). Indeed, due to the convention that the root of a positive real number is a positive real number, the form of the right-hand side implies that a choice was made.
(*) in the sense that, when raised to the fourth power, it gives $-2$.
