# 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.

• What's wrong with $x=1$? – Simply Beautiful Art May 31 '17 at 11:48
• @SimplyBeautifulArt: the same as $\sqrt x=-1$, which is impossible with the ordinary definition in the reals. – Yves Daoust May 31 '17 at 12:05
• @SimplyBeautifulArt The stumbling point here is that $x=1$ does not verify the original equation $x^{1/4} = -1$ unless one agrees to understand that $z \mapsto z^{1/4}$ uses non-principal branch. This point confused the OP, and needs to be clearly stated. – Sasha May 31 '17 at 12:45

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 $z=1$.

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.)

• The equation is precisely what the OP provided as her solution for the auxiliary equation resulting after taking the logarithm of both sides. This operation introduced extraneous roots, of which we must select those that satisfy the original equation, and there are none, unless $x^{1/4}$ refers to a non-principal branch. I think this answer just confuses the OP (-1). – Sasha May 31 '17 at 12:18
• As I say, the OP is having a problem with $x=1$, because it does not verify the original equation, unless one uses non-principal branch of the fourth root to evaluate $(1)^{1/4}$ to $-1$. – Sasha May 31 '17 at 12:42
• The root $x=1$ is extraneous for the primary branch of the root $x^{1/4}$. – Sasha May 31 '17 at 13:58
• By converting the original equation to $\arg x$, you are solving a different equation, i.e. for real $\varphi$, solve $\exp\left(i \varphi/4 \right) = -1$, which you have solved to $\varphi = 4 \pi + 8 \pi k$, then $z = \exp\left(i \varphi\right)$. Note however, that you have implicitly assumed $\left(\mathrm{e}^{i \varphi}\right)^{1/4} = \mathrm{e}^{i \varphi/4}$ which is equivalent to adopting the principal branch of $z^{1/4}$. Substituting the solution for $\varphi$ into $z = \exp\left(i \varphi\right)$ gives $z=1$, which is extraneous, as $(1)^{1/4} = 1$ on the principal branch. – Sasha May 31 '17 at 14:41