Just out of curiosity, I would like to know if this derivation is correct or not.

Let's assume complex numbers and write $1 = e^{2\pi i n}$, for any $n\in\mathbb{Z}$.

Then, by exponentiation we obtain $$1^{\frac{-i\ln 2}{2\pi}}=e^{2\pi i n \cdot \frac{-i \ln 2}{2\pi}} = 2^n,$$ and thus for $n=1$, $1=2$.

For me, this looks like a big contradiction. Any power of $1$ should be equal to $1$, or? What is the catch here that I don't see? In complex numbers, the power of $1$ doesn't have to be equal to $1$? Thanks.

  • $\begingroup$ You can take negative $n$ too, the possible values of $1^{\frac{-i\ln 2}{2\pi}}$ are $2^n,\; n\in \mathbb{Z}$. Which one you get depends on the used branch of the logarithm. $\endgroup$ – Daniel Fischer May 12 '15 at 21:45
  • $\begingroup$ @DanielFischer Fixed. Thanks. $\endgroup$ – pisoir May 12 '15 at 21:51

The rule $(a^b)^c=a^{bc}$ does not hold for complex numbers.

  • 3
    $\begingroup$ Or it works, but only when treating exponentiation as a multi-valued function. $\endgroup$ – Thomas Andrews May 12 '15 at 21:49
  • $\begingroup$ @ThomasAndrews Could you please give some example? To understand better the difference between "normal" exponentiation and exponentiation as a multi-valued function. $\endgroup$ – pisoir May 12 '15 at 22:00
  • $\begingroup$ It all comes down to the fact that $\log(z)$ is not well defined for $z$ a complex number, there are infinitely many values $w$ such that $e^w=z$. Thus since $a^b=e^{b\ln(a)}$ (that's how $a^b$ is defined for complex numbers), $a^b$ is not uniquely defined. You should read the first couple chapters of a complex analysis book. $\endgroup$ – Gregory Grant May 12 '15 at 22:03
  • $\begingroup$ @GregoryGrant From your answer I understand that since it is not uniquely defined, the results 1=2 in complex numbers (when I pick non-principal branch of the logarithm) is valid. It is the same as if I write $e^{-i\frac{\ln 2}{2\pi}\ln(1)}=e^{-i\frac{\ln 2}{2\pi}(\ln|1|+i2\pi n)}=2^n$. $\endgroup$ – pisoir May 14 '15 at 16:35

For $k \in \mathbb{Z}$ and $z \in \mathbb{C}$,

$$1^z = (e^{i2\pi k})^z = e^{i2\pi k \cdot z}$$

If $z = x$ where $x \in \mathbb{R}$, then $1^z = e^{i2\pi k \cdot x}$ can be any number on on the complex unit circle. Note that the $x$ is rotating the point $z=1$ around the circle.

If $z = x + iy $ where $x,y \in \mathbb{R}$, then $1^z = e^{-y}e^{i2\pi k \cdot x}$ can be any number on the complex plane except zero. Again we have a rotation by $x$. But now the magnitude can assume any non-vanishing value, depending on $y$.

In a nutshell, $1^z$ does not always equal 1 when dealing with complex numbers. Spooky! You may be interested in trying to find what sort of $z$ gives $1^z = 1$.

  • $\begingroup$ What you are saying is that in complex numbers, since the exponentiation is not uniquely defined, the powers of $1$ can be anything. Nice. I never realized that before:) $\endgroup$ – pisoir May 14 '15 at 16:42

$$1^{\frac{-i\ln 2}{2\pi}}=$$

$$\left|1^{\frac{-i\ln 2}{2\pi}}\right|e^{\arg\left(1^{\frac{-i\ln 2}{2\pi}}\right)i}=$$

$$|1|^{\frac{-i\ln 2}{2\pi}}e^{\arg\left(1^{\frac{\ln(2)}{2\pi}e^{-\frac{1}{2}\pi i}}\right)i}=$$






  • $\begingroup$ You only considered principal branch for $n=0$. What about other options? $\endgroup$ – pisoir May 14 '15 at 16:38

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