# What is wrong with this fake proof $e^i = 1$?

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

Obviously, one of my algebraic manipulations is not valid.

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The second equality: $(a^b)^c$ is not equal to $a^{bc}$ when $a,b,c\in\mathbb{C}$. Both expressions are multivalued functions, i.e. they can have several meanings. – Stefan Hansen Jan 18 '13 at 17:49
@StefanHansen, you can elaborate (not because it is insufficient but to make it worthy of being called answer.) a little and answer it . – 007resu Jan 18 '13 at 18:15
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... – N. S. Jan 18 '13 at 19:08
Or $1=1^{1/2}=(e^{i 2 \pi })^{1/2}= e^{i \pi} = -1$ – leonbloy Jan 18 '13 at 19:11

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.