5
$\begingroup$

Why doesn't $e^x$ have an inverse in the complex plane? Can someone please clarify it?

$\endgroup$
  • 5
    $\begingroup$ Hint: $e^{2\pi i} =e^{0}$... $\endgroup$ – Fabian Oct 11 '15 at 19:34
5
$\begingroup$

Among reals, only $0$ has the property that $e^0 = 1$, but among complex numbers, there are many $z$ such that $e^z=1$, for example, $2\pi i$, $4\pi i$, $6\pi i$ etc. But since $e^{z+w} = e^z*e^w$, you could add any of those numbers to any exponent $w$ and the value of $e^w$ doesn't change. Therefore $e^w$ is not one-to-one and so cannot have an inverse.

$\endgroup$
4
$\begingroup$

Here comes the overkill: by Great Picard's Theorem, any analytic function with an essential singularity at infinity takes every complex value, with at most one exception, an infinite number of times. $e^z$ clearly has an essential singularity at infinity.

$\endgroup$
  • 1
    $\begingroup$ That is most certainly not true. How about $f(z)=z$? (You're quoting big Picard, not little Picard, so you want a non-polynomial to get an essential singularity at $\infty$.) $\endgroup$ – mrf Oct 11 '15 at 19:45
  • 1
    $\begingroup$ @mrf: many thanks, now fixed. $\endgroup$ – Jack D'Aurizio Oct 11 '15 at 19:54

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.