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

  • $\begingroup$ The simple answer is that there are three possible cube roots, two possible square roots, and four possible fourth roots. With complex numbers it is not obvious which to select $\endgroup$
    – Henry
    Commented Jan 9, 2017 at 0:12
  • $\begingroup$ You should search the site for "complex exponentiation". The real fact is that $(a^b)^c=a^{bc}$ does not work in the complex plane because the log function is mutivalued, but we shouldn't have to rewrite that. $\endgroup$ Commented Jan 9, 2017 at 0:13
  • $\begingroup$ I know the multi values thing but i didn't know it was the reason for this i'll do my searches thank you $\endgroup$ Commented Jan 9, 2017 at 0:18

2 Answers 2


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.


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.

  • $\begingroup$ Under almost all contexts, I do not think we would have $\sqrt1=-1$. On the contrary, I think it is standard to use $1^{1/2}=-1$ or $x^2=1\implies x\stackrel?=-1$. $\endgroup$ Commented Jan 9, 2017 at 0:23
  • $\begingroup$ @SimpleArt you're right, I was using it as a means to show the problem with multivalued functions. $\endgroup$
    – Hugh
    Commented Jan 9, 2017 at 0:25
  • $\begingroup$ Thank you for your answer but again something comes to mind... according to the multivalues is $(-1)^{\frac {1}{2}} = i or -i $? $\endgroup$ Commented Jan 9, 2017 at 0:30
  • $\begingroup$ @magedrifaat Depends how you define it. That is, which branch are you taking it? $\endgroup$ Commented Jan 9, 2017 at 0:33
  • $\begingroup$ @magedrifaat if your function is multivalued then it's both $i$ and $-i$ $\endgroup$
    – Hugh
    Commented Jan 9, 2017 at 1:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .