Is ${(-1)}^{\frac{1}{6}}=i$?

Question

I just thought of this and tried to simplify. ${(-1)}^\frac{1}{6}$.

$${(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}$$

Since the cube root of $-1$ is $-1$,

$${(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i$$

Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?

Well, $i^6=-1$, but there are other complex numbers (including $-i$) with that property as well; Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when m and n are not integers. — J. W. Tanner, Mar 10 '20 at 03:12
@J.W.Tanner but does that mean ${(-1)}^{\frac{1}{6}}$ is indeed $i$? — harpey1111, Mar 10 '20 at 03:20

Doug M · Answer 1 · 2020-03-10T03:32:00.383

1

If $z = (-1)^\frac 16$ then $z^6 = -1$ or $z^{6} + 1 = 0$

This we can factor!

$(z^{2} + 1)(z^4 - z^2 + 1) = 0$

From the left hand factor, $z = \pm i$ are indeed solutions.

The right hand factor is a little trickier.

$z^4 - z^2 + 1 = (z^2 + \sqrt 3 z + 1)(z^2 - \sqrt 3z + 1)$

$z = \pm\frac {\sqrt {3}}{2} \pm \frac {1}{2} i$

There are 6 solutions.

These are called the roots of unity.

There are other methods to find these roots.

If you know De Moivre's law you can apply it.

$z^n = |z|^n(\cos \theta + i\sin \theta)^n = |z|^n(\cos n\theta + i\sin n\theta)$

Using this approach:

$(-1)^\frac 16 = (\cos \pi + i\sin \pi)^\frac 16 = \cos \frac {\pi}{6} + i\sin \frac {\pi}6$

$\cos \frac {(2k-1)\pi}{6} + i\sin \frac {(2k-1)\pi}6$ are also roots.

And by Euler formula

$e^{\pi i} = -1\\ e^{\frac {\pi}{6} i} = (-1)^\frac 16$

edited Mar 10 '20 at 03:32

answered Mar 10 '20 at 03:26

Doug M

I also thought of the Principal Square Root. This means the positive square root of a number is given when simplifying just a square root equation. But if it's equation, we use both the positive and negative. Does that apply to any even roots? My example is just an expression simplified, so can consider the Principal root to be the answer since it's just an expression simplified? – harpey1111 Mar 10 '20 at 03:34
With complex roots, the principal root would be the one with the smallest value of $\theta.$ i.e. $\frac {1}{2} +\frac {\sqrt 3}{2} i$ – Doug M Mar 10 '20 at 03:36

score 0 · Answer 2 · answered Mar 10 '20 at 03:21

It is true that $i^6=-1$, but fractional powers of negative numbers are not uniquely defined,

and the "general rule" $(a^m)^n=a^{m×n}$ does not always work when $m$ and $n$ are not integers.

There are six complex $6^{th}$ roots of $-1$; they are $\pm i$ and $\pm\dfrac{\sqrt3} 2\pm \dfrac i2$.

2 Answers2