Why can't we separate fraction powers in euler's formula?

Question

I am sure there is something wrong i am doing but i just want someone to point it to me.

Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$ Why do we have to calculate it by the formula as it is (which produces different value)

Another question is why can't we say that $ (-1)^{\frac {1}{2}} = \left(\left (-1\right)^{2}\right)^{\frac {1}{4}} = 1^{\frac {1}{4}} = 1$

Again i have to say that i know this is all wrong i just want to know why it is wrong

Answer 1

It is simply the case that $(a^b)^c$ is not always equal to $a^{bc}$ if $a$ is not positive or if $b$ and/or $c$ are complex.

The problems here are that

$$x=(-1)^{1/2}\implies x^2=-1\implies x^4=1$$

However, when you say that $1^{1/4}=1$, you cause a misconception. Usually, this is perfectly fine, but when you think about it, why not $1^{1/4}=-1$? Indeed, both values are solutions to the equation $x^4=1$, but neither are equal to $(-1)^{1/2}$. However, upon factoring, you could see that

$$x^4=1\implies x^4-1=0\implies(x+1)(x-1)(x+i)(x-i)=0$$

One of these is the correct solution in our context, though you happen to choose the wrong one.

Similarly, you state that $(-1)^{1/3}=-1$, but

$$x^3=-1\implies x^3+1=0\implies(x+1)(x^2-x+1)=0$$

And $x=e^{\pi i/3}$ is one such possible solution. If you want more information, please see the link above.

Answer 2

The laws we learn for powers apply to positive real numbers, they don't apply when you start to take roots of negative numbers.

Your proof that $ (-1)^{\frac {1}{2}}=1$ obviously raises alarm. As Ross mentioned in the comments this is related to functions being multivalued.

A similar false-proof using multivalued functions is $\sqrt{1}=1$ but $\sqrt{1}=-1$. Therefore $1=-1$.

Because we allowed the square root to have two results it lead to the result $1=-1$. Normally the square root is defined to be always positive so as to avoid this problem. However, if we allow for multiple values then we have to remember that $\sqrt{a}=b$ and $\sqrt{a}=c$ doesn't imply $b=\sqrt{a}=c$. Allowing multivalued functions to exist means that we lose a property of equality.