what is (-1)^(2/3)

Question

Google says that ${(-1)}^{2/3}$ is $-0.5+\frac{\sqrt{3}}{2}i$ but on socratic it says that it is 1. Which one is it?

Answer 1

$(-1)^{2/3}=[(-1)^2]^{1/3}=\sqrt[3]{1}$.

Now, if you work ONLY on real numbers, the result is ONLY 1.

If you work on COMPLEX numbers, there are 3 cube roots of unity: $1,\, -\frac{1}{2}\pm\frac{\sqrt3}{2}i$.

To get them, let write $1$ in polar form ($r_\theta$, where $r$ is the modulus and $\theta$ is an argument): $1=1_0$.

So $\sqrt[3]{1_0}=r_\theta \iff 1_0=(r_\theta)^3=(r^3)_{3\theta}$. Hence, $r^3=1$, whence $r=1$; and $3\theta=0+2k\pi$ with $k\in\mathbb Z$. So, $\theta=\frac{2k\pi}{3}$ with $k\in\mathbb Z$, but it is enough to take $k=0,1,2$.

So, all cube roots are:

• For $k=0$ $\Rightarrow$ $1_0=1$.
• For $k=1$ $\Rightarrow$ $1_{2\pi/3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$.
• For $k=2$ $\Rightarrow$ $1_{4\pi/3}=-\frac{1}{2}-\frac{\sqrt3}{2}i$.

Answer 2

Among the complex numbers, there are many ways to express the number $-1$. In particular, $-1=e^{(1 + 2n) \pi i}$, where $n \in \mathbb{Z}$. However, when substituting these in the equation $$ x = (-1)^{2/3}, $$ these cases all "boil down" to $3$ cases: $$\begin{align} x&= e^{2 \pi i} = 1, \\ x&= e^{\frac{2}{3} \pi i} = -0.5 + i\ 0.866025\ldots,\\ x&= e^{-\frac{2}{3} \pi i}= -0.5 - i\ 0.866025\ldots, \end{align} $$ which are the three solutions of the equation above.