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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

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  • $\begingroup$ 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 Jan 9 '17 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. $\endgroup$ – Ross Millikan Jan 9 '17 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 $\endgroup$ – maged rifaat Jan 9 '17 at 0:18
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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.

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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.

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  • $\begingroup$ 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 $x^2=1\implies x\stackrel?=-1$. $\endgroup$ – Simply Beautiful Art Jan 9 '17 at 0:23
  • $\begingroup$ @SimpleArt you're right, I was using it as a means to show the problem with multivalued functions. $\endgroup$ – Hugh Jan 9 '17 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 $? $\endgroup$ – maged rifaat Jan 9 '17 at 0:30
  • $\begingroup$ @magedrifaat Depends how you define it. That is, which branch are you taking it? $\endgroup$ – Simply Beautiful Art Jan 9 '17 at 0:33
  • $\begingroup$ @magedrifaat if your function is multivalued then it's both $i$ and $-i$ $\endgroup$ – Hugh Jan 9 '17 at 1:40

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