I have had this problem for a long time now:

Suppose we take the cube root of unity $\omega$.

$\omega ^ 2 = e ^ {i {4\pi\over3}} = {(e ^ {i 4\pi})}^{1\over3} = {(e ^ {i 2\pi})}^{1\over3} = e ^ {i {2\pi\over3}} = \omega$

Why doesn't this manipulation work? Are we not allowed to divide exponents like we do with real numbers? Could anyone explain the reasons clearly?Could this be similar to the square root function, where the result should always be positive?

Also, how does the square root function hold in the realm of complex numbers? How do we decide the positive output here? For example, is $\sqrt{3+4i}$ equal to $2+i$ or can it be $-2-i$ too? More importantly, can $\sqrt{-1}$ be written as $-i$ or is it defined to be only $i$?

  • $\begingroup$ For complex $a,x,y$ you normally define $(a^x)^y=e^{y\ln(a^x)}.$ The problems come basically from the failure of the law $\ln(a^x)=x \ln(a)$, which is valid for real $a,x > 0$ but not necessarily for other real or complex $a,x.$ $\endgroup$ – gammatester Feb 8 '16 at 14:02

$e^{n\pi i}=\cos(n \pi) + i \sin(n\pi)$.

In your case you are claiming:

$\cos(\frac{4\pi}{3})+i\sin(\frac{4\pi}{3})=(\cos(4\pi)+i\sin(4\pi))^{1/3}$ which is a contradiction.

  • $\begingroup$ I know that it is a contradiction. I want to know what rules I flouted that made this contradiction possible. $\endgroup$ – user117913 Feb 8 '16 at 14:03
  • $\begingroup$ Logarithms and square roots are not globally defined as holomorphic functions so in general we can't take nth root of something. $\endgroup$ – Abellan Feb 8 '16 at 14:04
  • $\begingroup$ I'm not sure I understand what you mean by that $\endgroup$ – user117913 Feb 8 '16 at 14:05
  • 1
    $\begingroup$ You should read the wikipedia entry on complex logarithm if you really want a true answer. Also exponentials in complex numbers don't work as we're used to just think that in you question you are claiming $\omega=1$ since $e^{2 \pi i}=1$ $\endgroup$ – Abellan Feb 8 '16 at 14:10
  • $\begingroup$ That's the point I'm trying to make here. I don't claim that to be true. I want to know why it cannot claimed to be true, specifically, which boundaries I crossed in this manipulation. $\endgroup$ – user117913 Feb 8 '16 at 14:16

It's not necessary that root should always be positive. If it was root of 3-4i then -2-2i is correct and yes i(iota) is defined specifically as root of -1 which is imaginary


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.