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

What's wrong with this proof? I'm guessing that the flaw is involved with the 3rd equal sign, where $e^{\frac{2\pi i}{2\pi i}} = (e^{2\pi i})^{\frac{1}{2\pi i}}$. But what is the specific rule that's being broken? More importantly, what is the rule I need to follow generally in complex analysis to make sure this kind of thing doesn't happen?

  • 4
    $\begingroup$ For complex numbers, you can't do $(x^y)^z=x^{yz}$. $\endgroup$ – Thomas Andrews Sep 5 '17 at 18:40
  • 3
    $\begingroup$ This is like saying $-1 = (-1)^{1}=(-1)^{2/2}=((-1)^2)^{1/2}=1^{1/2}=1$. $\endgroup$ – Thomas Andrews Sep 5 '17 at 18:41
  • $\begingroup$ $a^{bc} = (a^b)^c$ isn't always true for complex numbers $\endgroup$ – Zonko Sep 5 '17 at 18:41
  • $\begingroup$ $e^1 = e^{1+2 \pi k i}$ with $k \in \mathbb{Z}$ $\endgroup$ – Blex Sep 5 '17 at 18:41
  • $\begingroup$ The fifth equality is also not valid. $1^z$ can be any of the values $e^{2\pi k i z}$ for integral $k$, not just the particular value $1$ you get by taking $k=0$. $\endgroup$ – MPW Sep 5 '17 at 18:47

This is like saying:


The problem is that complex $x^{y}$ is actually tricky.

If we want $x^y$ to be a single-valued function, then you lose the property $\left(x^y\right)^z=x^{yz}$.

If we allow $x^y$ to be multi-valued, then $-1$ and $1$ are both values of $1^{1/2}$, and $e$ and $1$ are both values of $1^{\frac{1}{2\pi i}}$. But that doesn't mean that $-1=1$ or $e=1$.

In generally, if $x$ is a non-zero complex number, the only time there is a single value for $x^y$ is when $y$ is an integer.

When $y=p/q$ is a rational number, with $p,q$ relatively prime, then there are $q$ distinct possible values for $x^y$.

For any other $y$, there are infinitely many possible values of $x^y$. In particular, there are infinitely many values for $1^{\frac{1}{2\pi i}}$, and they are all $e^{k}$ for some integer $k$.

With the multivalued view you only get that every value of $x^{yz}$ is a value of $(x^y)^z.$ The opposite is not true, in general. Similarly, every value of $x^{y+z}$ is a value of $x^yx^z,$ but not visa versa.


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