Solving equations with complex numbers and rational power

Question

How to solve equation $x^{\frac{3}{2}}=i$? De Moivre's formula fails. Wolframalpha gives $x=\frac{1+\sqrt{3}i}{2}$.

Is there any other solutions? Is there infinitely many solutions?

Answer 1

$x^{\frac 32} = k$ means that i) $x^3 = k^2$ and that $-\frac \pi 2 < \arg k \le \frac \pi 2$.

So $x^3 = i^2 = -1$ and so by De Moivre $x= \frac {1 +\sqrt 3}2, \frac {1-\sqrt 3}2,-1$. Of these three, the only one with an argument between $-\frac \pi 2$ and $\frac \pi 2$ is $x= \frac {1 +\sqrt 3}2$

==== old incorrect answer ====

What do you mean De Moivres formula fails?

$i = e^{i(\frac \pi2 + 2k\pi)}=\cos \frac \pi 2 + i \sin \frac \pi 2$. so $x^{\frac 32} = i$ is solved by $e^{i(\frac 23(\frac \pi 2 + 2k \pi))}=\cos (\frac 23(\frac \pi 2 + 2k \pi)) + i \sin (\frac 23(\frac \pi 2 + 2k \pi))$

Note if $k=0$ then $x= \cos \frac \pi 3 + i\sin \frac \pi 3 = \frac 12 + i\frac {\sqrt 3}2$ which is wolfram's answer.

==== to make the old answer correct we must add....====

However $x$ must be a principal square root so we must have $-\frac \pi 2 < \frac 23(\frac \pi 2 + 2k \pi) \le \frac \pi 2$.