# Extraneous solutions to $i^{2/3}$

I want to find the value of $$i^{2/3}$$

Here was what I tried: $$i^{2/3} = (i^{2})^{1/3} = -1^{1/3} = (-1^{2})^{1/6} = 1$$

I know that I could have also stopped at the third step, since

$$-1^{1/3} = -1$$

Clearly there are multiple solutions, and I was wondering if there was a good way of conceptualizing these multiple solutions. How many solutions does a complex number to the $$2/3$$ power have, and can I use the complex plane to visualize this?

• Welcome to Mathematics Stack Exchange. A non-zero complex number has three cube roots. If one is $\zeta$, the other two are $\zeta\omega$ and $\zeta\omega^2$, where $\omega$ is a complex cube root of $1$ – J. W. Tanner Sep 21 at 3:15
• To visualize, draw the circle $|z|=1$, locate the point $-1$ on this circle, find two other points on the circle that form an equilateral triangle with $-1$. Those two points, together with $-1$, are the three complex values of $(-1)^{1/3}$ (and of $i^{2/3}$). – Gerry Myerson Sep 21 at 3:15
• @GerryMyerson this is the sort of answer I was looking for in terms of graphical visualization. If I wanted to find the fourth root of a number could I draw a square / fifth root a normal pentagon? Is there an intuitive explanation of why drawing these shapes would correspond to roots (assuming what I said in the previous sentence is true)? – James McGreivy Sep 21 at 4:14
• Yes, what you say is true. Raising a complex number to a (whole number) power $n$ has the effect of multiplying its argument (the angle it makes with the real axis) by $n$, and the points that line up with the real axis when you multiply that angle by $n$ are the $n$ vertices of a regular $n$-gon. E.g., the points one-fifth, two-fifths, three-fifths, and four-fifths of the way around all align with the real axis when you multiply the angle by five. So that's why the $n$th roots of a number lie at the vertices of a regular $n$-gon. – Gerry Myerson Sep 21 at 10:25
• So, are we all set now, James? – Gerry Myerson Sep 22 at 23:55

In complex analysis, $$a^b$$ is defined by $$e^{b\log a}$$, where the complex log, i.e., $$\log a$$, is multi-valued and given by $$\log|a|+i (\arg(a)+2\pi k)$$ ($$k\in \mathbb{Z}$$) (in $$\log|a|$$, we use the usual real log). In your case, we have $$i^{2/3}=e^{2/3\cdot(\log|i|+i(\arg i+2\pi k))} = e^{\pi i/3+4k\pi i/3}\,,$$ which takes three different values.
Alternatively, you could use the fact that $$i = \cos (\pi/2)+ i \sin(\pi/2)$$ to evaluate the power (https://en.wikipedia.org/wiki/Complex_number): $$i^{2/3} = \cos \left(\frac{\pi}{3} + \frac{4k\pi }{3}\right)+ i \sin\left(\frac{\pi}{3} + \frac{4k\pi}{3}\right), \\ \text{where}\ k = 0, 1, 2$$