Does every analytic function pass through all points in the complex plane? If a function $f(z)$ is analytic and defined everywhere, do its values pass through every point in the complex plane?
In other words, is there for any $z$ in the complex plane an $a$ in $\mathbb{C}$ such that $f(a)=z$?
 A: 
In other words, is there for any $z$ in the complex plane an $a$ in $\Bbb C$ such that $f(a)=z$?

No.  Two counter-examples:

*

*A constant function takes only 1 value.


*$\exp(z) \neq 0$ for any $z\in \Bbb C$, same for the Gamma function.
There are however results like Picars's Theorems (cited from Wikipedia):

Little Picard Theorem: If a function $f : \Bbb C \to \Bbb C$ is entire and non-constant, then the set of values that $f(z)$ assumes is either the whole complex plane or the plane minus a single point.

and

Great Picard's Theorem: If an analytic function $f$ has an essential singularity at a point $w$, then on any punctured neighborhood of $w$, $f(z)$ takes on all possible complex values, with at most a single exception, infinitely often.

Notice that in the latter case allows $w=\infty$.  For example, $f(z)=\exp(1/z)$ has an essantial singularity at 0, this in any neighborhood of 0, $f$ takes all complex number with at most one exception.  For this $f$, the is actually one exception, namely $f(z) = 0$ is never realized.
The theorem also applies to $f(1/z)$ which has an essential singularity at $\infty$, i.e. in any neighborhood of $\infty$, $e^z$ takes on each complex number as a value infinitely many times, with the one exception that $e^z\neq 0$.
