$e^z, \ \ z \in \mathbb{C}$ isn't invertible Could you tell me why the function $\mathbb{R}^2 \ni z \rightarrow e^z \in \mathbb{R}^2$, complex exponential, is not invertible globally. On a horizontal strip $[\ i y, \ i(y+2 \pi))$ its inverse is $\ln z$.
I found this example on this forum, but there is no explanation if this fact. 
Or at least it is not explanatory enough for me to understand.
I would really appreciate all your insight.
Thank you.
 A: Periodic functions are not invertible, and $e^z$ is $2\pi i$-periodic.
A: I think you have to understand that a general function $f : A\to B$ to be invertible need first to be injective. Why that? Well, because recall for each $x\in A$ there must be unique $y\in B$ with $y=f(x)$. If $f$ is not injective, then there are $x_1,x_2\in A$, $x_1\neq x_2$ with $f(x_1)=f(x_2)$. Now, try defining the inverse, you can easily see that if $g : B \to A$ must be such that $g(f(x))=x$, then you have a problem, because you would have $g(f(x_1))=x_1$ and at the same time $g(f(x_2))=x_2$, but $f(x_1)=f(x_2)=y$, so for this $y$ there are two values in $A$, so $g$ isn't a function.
If a function $f: \Bbb C\to \Bbb C$ is periodic, that is there is $k\in \Bbb C$ with $f(z+k)=f(z)$, then it's clearly not injective, hence not invertible. As you can see
$$\exp(z+2\pi i)=\exp(x+iy+2\pi i)=e^x(\cos (y+2\pi) + i\sin(\theta+2\pi))=\exp(z),$$
so it can't be invertible. However all is not lost, when a function is not injective, you can restrict it's domain so that it becomes injective. You simply take away from it's domain the points on which it "duplicates". What this means in this case?
This means that since the function repeats when you add $2\pi i$ to the point, in every horizontal strip of vertical length less than $2\pi$ the function doesn't have duplicate values: it's injective, hence invertible. More precisely, let $\alpha \in \mathbb{R}$, then you let $S_{\alpha}=\{z \in \mathbb{C} : \alpha < \Im(z) < \alpha + 2\pi\}$, in this region $\exp$ is injective and invertible.
