Why is $(-3)^\frac{2}{5}$ not real?

Question

Solving some past year school questions, I noticed this simple exponentiation question. $$ \text{True/False?} \\ ((-3)^2)^\frac{2}{5} = ((-3)^\frac{2}{5})^2 $$

At first, this seemed to be right, however, the answer claims it's incorrect. It says RHS is not real. I suppose the logic is this: $$ (-3)^\frac{2}{5} = x \implies x^\frac{5}{2} = -3 \implies (x^5)^\frac{1}{2} = -3 \implies \sqrt{x^5} = -3 $$ Which is a contradiction. But I don't agree, since the same argument can be applied to $(-3)^\frac{2}{1}$.

Now as far as I understand, $(-3)^\frac{2}{5} = ((-3)^\frac{1}{5})^2 = ((-3)^2)^\frac{1}{5} = \sqrt[5]{9}$ Which is real. But also, on the other hand, my calculator claims it's $0.479546655 + 1.47589285 i$, which I suspect is treating the expression as: $$ (-1)^\frac{2}{5} \times 3^\frac{2}{5} = (e^{i\pi})^\frac{2}{5} \times 3^\frac{2}{5} = e^{i\frac{2\pi}{5}} \times 3^\frac{2}{5} = (\cos(\frac{2\pi}{5}) + i.\sin(\frac{2\pi}{5})) \times 3^\frac{2}{5} $$ which is complex. Is there a concrete answer to this question? Or does it eventually come down to the author's definition of fractional exponentiation?

Answer 1

It seems you have to find a definition what a^b is supposed to be if b is a fraction and a < 0.

For example $x^{1/3}$ should be defined for all real x but $x^{1/4}$ should require x >= 0. I think you an define $a^b$ for most rational b if you put b into minimal terms and you don’t have $a^{p/q}$ with q even and a < 0. For your question, 5th root of 9 would be fine.