Is the 6th root of -1 equals i (imaginary number i)?

[onesun0000]

I just thought of this and tried to simplify. [MATH]{(-1)}^\frac{1}{6}[/MATH]
[MATH]{(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}[/MATH]
Since the cube root of [MATH]-1[/MATH] is [MATH]-1[/MATH]
[MATH]{(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i[/MATH]
Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?

 

[Romsek]

[MATH]-1 = e^{i \pi(2n+1)},~n \in \mathbb{Z}\\ (-1)^{1/6} = e^{i \pi \frac{2n+1}{6}},~n = 0 , 1, \dots, 5\\ \text{Thus the 6 roots of -1 are }\\ e^{i\pi/6},~e^{i\pi/3},~e^{i\pi/2} = i,~e^{i5\pi/6},~e^{i7\pi/6},~e^{i3\pi/2}=-i,~e^{i11\pi/6} [/MATH]

 

[onesun0000]

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?

 

[Romsek]

[MATH]x^{2k} = c\\ (x^2)^k = c\\ x^2 = \sqrt[k]{c}\\ x = \pm \left(\sqrt[k]{c}\right)^{1/2}[/MATH]

 

[HallsofIvy]

As Romsek said there are 6 sixth roots of a complex number. We don't normally talk about the root of a non-real number. We do define the principal root as the root with smallest argument. With an odd root that smallest argument is 0 and the principal root is a positive real number. Here, with an even root, the smallest argument is \(\displaystyle \frac{\pi}{6}\) and the principal sixth root is \(\displaystyle e^{\frac{i\pi}{6}}\).