If $\omega^3 =1$ then how is $\omega^5=(\omega^{3})^{5/3}=1$?

Question

While revising the chapter complex number, when i was reading the the concept of the cube roots of unity, I read $\omega^3=1$, where $\omega$ is the cube root of unity other than 1, then $\omega ^5=\omega^3.\omega^2=\omega^2(\because\omega^3=1)$

the above thing is we all know, is correct, but you see i don't know how my brain functioned and i thought that $$\omega^5=(\omega^3)^{\frac 53}=(1)^\frac 53=1 \{\because \omega^3=1\}$$

but you know, this is wrong, i don't know how this wrong but i can see this to be wrong because if this true then $\omega ^k=1, k\in R$.

So how $$\omega^5=(\omega^3)^{\frac 53}=(1)^\frac 53=1 $$ is wrong? what are the wrong assumptions which is taken by me while writing that statement?

Answer 1

The wrong assumption lies in the step where you write:

$$\omega^5=(\omega^3)^\frac{5}{3}$$

This is wrong since it assumes the existence of a mapping $u$ similar to the power $t↦t^\frac35$ on $\mathbb{R}^+$ in the sense that $u(z^5)=z^3$ for every $z$ in $\mathbb{C}$. Such a mapping $u$ does not exist eventhough $u$ does exist on $\mathbb{R}$.

Answer 2

This is of course wrong. The rule you forgot is the use of rational exponents is valid only for real positive numbers.

An illuminating example is the following: consider the cube root of $-8$. You might want to write is as $(-8)^{1/3}=-2$.

However, since $1/3=2/6$, it would also be $\bigl((-8)^2\bigr)^{1/6}=64^{1/6}=+2$, or even $\bigl((-8)^{1/6}\bigr)^2$, and the sixth root of $-8$ doesn't even exist, in the real numbers. Of course, in the complex numbers, $-8$ has sixth roots – but it has six of them!

Here, the simplest thing to do is to use this fact: the non-real cubic roots of unity generate a cyclic group of order $3$, so $\;\omega^5=\omega^2\; (=\mkern1mu\overline{\mkern-1mu\omega\mkern-1mu}\mkern1mu=\omega^{-1})$.