Cube root of a complex number

Question

I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:

Show that $(-1 + \sqrt{3i})/2$ is a cube root of $1$.

The solution on linearalgebras.com shows the following solution here, at number 2.

It states that $(-1 + \sqrt{3i})/2$ squared is $(-1 - \sqrt{3i})/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?

Answer 1

The $i$ should not be inside the square root.

It should be $$\left(\frac{-1+\sqrt3 \,i}{2}\right)^{\!2}=\frac{-1-\sqrt3 \,i}{2}$$

rather than

$$\left(\frac{-1+\sqrt{3 i}}{2}\right)^{\!2}=\frac{-1-\sqrt{3 i}}{2}$$

It might be easier to understand it as $$\exp \left(\frac{2\pi i}3 \right)^{\!2}=\exp \left(\frac{4\pi i}3 \right)$$

Answer 2

Consider equation:

$x^3-1=0$

$$x^3-1=(x-1)(x^2+x+1)=0$$

$$x^2+x+1=0$$

$$x=\frac{-1±\sqrt {1-4}}{2}=\frac{-1±\sqrt 3 i}{2}$$