Solve equation $x^{1/4}=-1$ I tried 
$1/4\ln(x)=\ln(-1)+2ki\pi$
$\ln(x)=4(i\pi+4ki\pi)$
$x=1$
I must be wrong...
Math newbie here. Forgot how to do the complex algebras. Could you please give links or any help.
 A: Let's first agree on the domain where we are looking for solutions. We are solving the equation over complexes.
Complex function $z^{1/4}$ usually denotes the principal branch of solution $w(z)$ of $w^4 = z$, that is solutions with $-\frac{\pi}{4} < \arg\left(w\right) \leqslant \frac{\pi}{4}$.
For the principal branch, the equation $z^{1/4}=-1$ has no solution. 
The fourth root has three other branches, related to the principal branch as $\exp\left(i \pi k/4\right) z^{1/4}$ for $k=1$, $2$ or $3$. 
Using the branch corresponding to the choice $k=2$, where $w(z) = -\left(z\right)^{1/4}$, the equation $w(z) = -1$ has the solution of $z=1$.
A: In the reals, the equation has no solution. Similarly to the case of the square root, one must choose a branch of the fourth root and there is no reason to take another than the positive one. Hence $x^{1/4}\ge0$.
In the complex, taking the imaginary part of the logarithms,
$$\frac{\arg x}4=\pi+2k\pi,$$$$\arg x=4\pi+8k\pi$$ and $x=1$, which is the only solution. (Other branches give $x^{1/4}=1,i,-i$, but this is unimportant.)
