$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$

Obviously, one of my algebraic manipulations is not valid.

  • 3
    $\begingroup$ @StefanHansen, you can elaborate (not because it is insufficient but to make it worthy of being called answer.) a little and answer it . $\endgroup$
    – 007resu
    Commented Jan 18, 2013 at 18:15
  • 19
    $\begingroup$ The proof is exactly the same idea as $-1=\sqrt{(-1)^2}=\sqrt{1}=1$. It is less obvious just because the "root" part is the $\frac{1}{2\pi}$ power... $\endgroup$
    – N. S.
    Commented Jan 18, 2013 at 19:08
  • 15
    $\begingroup$ Or $1=1^{1/2}=(e^{i 2 \pi })^{1/2}= e^{i \pi} = -1$ $\endgroup$
    – leonbloy
    Commented Jan 18, 2013 at 19:11
  • $\begingroup$ @leonbloy When I first learnt of Euler's Identity, this was the first thing that came to mind. It is my favourite paradox (although not true) :D $\endgroup$
    – Mr Pie
    Commented Feb 27, 2018 at 14:17
  • $\begingroup$ As a complex power, the result of $1^{1/2\pi}$ is not $1$, but $e^{ki}$ where $k\in\Bbb Z\,.$ $\endgroup$
    – Angelo
    Commented Sep 12, 2023 at 5:57

2 Answers 2


As said in the comments, the expressions $(a^b)^c$ and $a^{bc}$ are in fact multivalued functions; they are not a uniquely determined complex number. A classic example of a multivalued function is the complex logarithm denoted by $\log(z)$, $z\in\mathbb{C}$. The complex logarithm $\log(z)$ is any complex number $w$ satisfying $e^w=z$ (which has several solutions, see e.g. this), and hence $\log(z)$ gives rise to a whole set of complex numbers instead of just a single complex number.

Complex exponentiation such as $z^w$ for $z,w\in\mathbb{C}$ is usually defined as $$ z^w=\exp(w\log(z)), $$ where $\log(z)$ is the complex logarithm, and hence this is also a multivalued function. I hope this sheds some light on the problems with doing manipulations on complex numbers as if they were real numbers. See also this for other examples of identities which fail when using complex numbers as they were real numbers.

  • $\begingroup$ Then proofs of demoivire theorum is wrong ? And also the derivation of cubes root of unity ? $\endgroup$
    – Get_ Maths
    Commented Aug 13, 2022 at 11:32

I'm going to add this here as the most general answer to this common problem in solving equations, a principle which the other answers do not cover:

When asked to solve for $y$, an equation of the form:


(in your example $f(x)=e^i$)

The approach commonly taken is to solve by writing a sequence of equivalent equations working towards $y=\text{the desired answer}$, in your case as follows:

$y=e^{i}\\ y= e^{i2\pi/2\pi} \\y= (e^{2\pi i})^{1/(2\pi)}\\y = 1^{1/(2\pi )}\\y = 1$

Then this method is only reliable if it is truthful to write if and only if between every consecutive pair of equivalent equations in sequence as follows:

$y=e^{i}\\\iff\\ y= e^{i2\pi/2\pi} \\\iff\\y= (e^{2\pi i})^{1/(2\pi)}\\\iff\\y = 1^{1/(2\pi )}\\\iff\\y = 1$

Due to the periodicity of complex powers, the $\iff$ statement is contradictory to the laws of arithmetic in various places in the above chain.


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