7
$\begingroup$

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

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?

$\endgroup$
  • 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
9
$\begingroup$

This is like saying:

$$-1=(-1)^{1}=\left((-1)^{2}\right)^{1/2}=1^{1/2}=1$$

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.

$\endgroup$

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.