Why the alternate form of $\frac{\sqrt3}2+\frac i 2$ is $\sqrt[6]{-1}$?

Question

The alternate form of: $$\frac{\sqrt3}2+\frac i 2$$ is $$\sqrt[6]{-1}$$ (I know that thanks to WolframAlpha.)

What are the arithmetic actions that gets us from the former to the latter?

Answer 1

That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $\sqrt[6]{-1}$.

The notation makes sense when we write $\sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.

The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=\cos\pi+i\sin \pi$), $$ \omega_k=\cos\left(\tfrac{\pi+2k\pi}{6}\right)+i\sin\left(\tfrac{\pi+2k\pi}{6}\right),\ \ k=0,\ldots,5. $$ Your root is $\omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $\omega_0^r=-1$.

Answer 2

$$\sqrt{3}/2+i/2=\cos(\pi/6)+i\sin(\pi/6)=e^{i\pi/6}"="(e^{i\pi})^{1/6}=(-1)^{1/6}.$$

The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{i\pi})^{1/6}$, and $e^{i\pi/6}$ happens to be one of them.

Answer 3

We simply have that

$$\left(\frac{\sqrt3}2+\frac i 2\right)^6=(e^{i\pi/6})^6=e^{i\pi}=-1$$

and since $\frac{\sqrt3}2+\frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $\sqrt[6]{-1}$.

Refer also to:

• Principal $n$th root of a complex number