Does i^(2/3) have two solutions

Per the exponentiation rules, i^(2/3) can be rewritten as (i^2)^(1/3) = (-1)^(1/3) = -1.

However, when I type in i^(2/3) on Google, I get this result: (0,5 + (0,866025404 * i)).

If i^(2/3) = -1, then (-1)^(3/2) is supposed to be i, but it's actually -i.

(0,5 + (0,866025404 * i))^(3/2) equals i, however (according to Google).

Are both solutions valid, or is there an exception regarding the exponential rules in this case?

..2/3
I.......has ONE VALUE and only one Value.

In fact

If x is a complex number

AND

If x is not equal to zero or one
t
hen

..a
x has a UNIQUE {one only} VALUE, where a is any real number.


Consider the simple real case of

..1/2
9.

That has a value of 3 only , although the
solution set for

x² = 9 is x=3 or x =-3

By convention

..1/2
x is the PRINCIPAL Square root of x only {the positive one when x is a positive real}



Now

..2/3
I.........does equal i² raised to the 1/3 power

and that is

.....1/3
(-1)

Guess what, THAT IS NOT EQUAL TO negative 1 .

The reason is that -1, which is a cube root of -1, IS NOT THE PRINCIPAL CUBE ROOT
over the complex number set.

The equation
.......3
x.......= -1

has three solutions in the complex numbers,

First solution is
............_
..1......√3
▬..+...▬ I which is 1 (cos 60° + I sin 60°)
..2.......2


Second solution is

.....-1.......... which is 1 (cos 180° + I sin180°)


Third solution is
............_
..1......√3
▬..-...▬ I which is 1 (cos 300° + I sin 300°)
..2.......2

THE PRINCIPAL SOLUTION IS ONLY THE FIRST ONE, -1 is the second one.


******
Note, if you did not understand

1( cos Θ + I sin Θ) or in greater generality