# when is exp(z) injective, surjective, bijective, holomorphic, biholomorphic

I would say holomorphic everywhere, injective and bijective only in $D=\mathbb{R}\times [-\pi,\pi]$, so also only biholomorph in that D.

Is that true? Are there alternatives?

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It fails to be injective (so to be bijective, too) in $\Bbb R\times[-\pi,\pi]$, actually. Consider for example $\exp(-i\pi)$ and $\exp(i\pi)$. Bear in mind that $\exp(z)$ has period $2\pi i$ (you may want to prove this as an exercise, if you've never seen this result). That should give you an idea how you can fix that issue to get injectivity back.
Now, if we're talking about surjectivity, we have to have a codomain specified. If your codomain is $\Bbb C\smallsetminus\{0\}$, then surjectivity can be obtained when the domain is a sufficiently large subset of $\Bbb C$--such as $\Bbb R\times[-\pi,\pi]$--but there is no $z$ such that $\exp(z)=0$, so we'll never get surjectivity if the codomain given is $\Bbb C$.
Once you've chosen an appropriate domain and codomain to get bijectivity, you are correct about $\exp(z)$ being biholomorphic.
its just that i read somewhere that the slit plane (which is the subset i gave) is good for something in relation to exp(z). Isnt what I wrote the same as $S=\{ w\in \mathbb{C} | -\pi < Im(w) < \pi \}$ –  bakabakabaka Sep 12 '12 at 20:08
Ah, I see. Actually, that's $\Bbb R\times(-\pi,\pi)$. What you had is the same as $\{w\in\Bbb C|-\pi\leq\text{Im}(w)\leq\pi\}$. –  Cameron Buie Sep 12 '12 at 20:24
It's also (perhaps) worth remarking that for any real $a$, the strip $\Bbb R\times(a-\pi,a+\pi)$ gets mapped bijectively by $\exp(z)$ to the plane with some slit removed from the origin to the point at $\infty$. In particular, If $a$ is any even multiple of $\pi$ (not just $0$), then it will be mapped to the plane with the nonnegative reals removed. –  Cameron Buie Sep 12 '12 at 20:29