# What does $\sqrt[6]{-1}$ mean on wolfram?

To solve $X^4-X^2+1=0$, wolfram wrote $\pm\sqrt[6]{-1}$ and $\pm(-1)^{5/6}$. Does it mean $\sqrt[6]{i}$ and $i^{5/6}$ or $\pm e^{\frac{i\pi}{6}}$ and $\pm e^{\frac{5i\pi}{6}}$ ?

In fact, nothing looks to be good since solution are $e^{\pm i\pi/6}$ and $e^{\pm 5i\pi/6}$, so why such notation ?

• $e^{i\pi/6}=\sqrt[6]{-1}$ since $(e^{i\pi/6})^6=e^{i\pi}=-1$. The solutions given by Wolfram are the same solutions as yours, just given in terms of roots of unity rather than by their exponential form. – Sean English Jul 10 '16 at 16:43

Wolfram means $$\pm\sqrt[6]{-1}=\pm\sqrt[6]{e^{i\pi}}=\pm e^{\frac{\pi i}6}$$ or in other words $$\pm\sqrt[3]{\sqrt[2]{-1}}=\pm\sqrt[3]{i}=\pm\sqrt[3] {e^{\frac{i\pi}{2}}}=\pm e^{\frac{\pi i}6},$$ as Wolframalpha takes the principal branch $(-\pi,\pi]$, and so $i=e^{\frac{i\pi}{2}}$ and $-1=e^{i\pi}$ when computing roots.

For the divergence of solutions, just note that $$-e^{\frac{i\pi}{6}}=e^{-\pi i}\cdot e^{\frac{i\pi}{6}}=e^{-\frac{5i\pi}{6}}$$ and so on. So the solutions are indeed the same.

• what do you mean by standard branch ? How can $i=e^{i\pi}$, whereas $e^{i\pi}=-1\neq i$ ? – MathBeginner Jul 10 '16 at 16:34
• Alpha / Mathematica does not take the branch $[0,2\pi)$. It uses $(-\pi,\pi]$. – Mario Carneiro Jul 10 '16 at 16:53
• absolutely, which is indeed the principal branch of $\log$ as I wanted to state. Terrible mistake by my part. – b00n heT Jul 10 '16 at 16:56

I assume the reason why Mathematica prefers $(-1)^{1/6}$ to $e^{i\pi/6}$ is because it has fewer transcendental constants, and is thus easier to work with (as a CAS). It's also simpler as a Mathematica expression, that is Power[-1, 1/6] as compared to Exp[Times[0 + 1/6 I, Pi]]; since the rational numbers are treated as atomic (as well as "complex rationals", i.e. elements of $\Bbb Q(i)$), that's one operation instead of two, which I'm sure is a factor in Simplify handling.

Mathematica uses the convention $a^b=e^{b\log a}$, where $\log$ has a branch cut at the negative reals. It follows from this definition that $(-1)^{1/2}=i$, $(-1)^{1/3}=e^{i\pi/3}$, and $(-1)^{1/6}=e^{i\pi/6}$ (and in general $(-1)^x=e^{i\pi x}$ for $x\in(-1/2,1/2]$).

The negative of $(-1)^{1/6}$ is $$-(-1)^{1/6}=e^{7i\pi/6}=e^{-5i\pi/6}=(-1)^{-5/6},$$ and the conjugate is $$\overline{(-1)^{1/6}}=e^{-i\pi/6}=e^{5i\pi/6}=(-1)^{5/6}.$$

Wolfram is right,

$$x^4-x^2+1=\frac{x^6+1}{x^2+1}$$ and the roots are those of $x^6+1$, but those of $x^2+1$.

But you are wrong to write $\sqrt[6]{-1}=\sqrt[6]{i}$ (as $-1\ne i$). Instead, $\sqrt[6]{-1}=\sqrt[3]{i}$, at least for some root of $-1$ and $i$.