# Why doesn't $e^x$ have an inverse in the complex plane?

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

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

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.
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.
• 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 $\infty$.) – mrf Oct 11 '15 at 19:45