If $f(z)$ is an entire analytic function assuming the values $0$ and $1$, show that for any complex number $a$ and any real number $ϵ>0$ there is a point $z_0$ such that $|f(z_0)−a|<ϵ$
My work:
Assume that there does not exist such a $z_0$ that fulfills the above inequality.
Then, given a complex number $a$, the function
$$h:= f(z)-a$$ does not have any roots.
By the fundamental theorem of algebra, $h$ cannot be a non-constant polynomial.
So we know that $f(z)$ must be an entire function that grows faster than a polynomial, and assumes the values 0 and 1, among other values (thanks to the comments posted below so far.).
Moreover, since the entire function is not constant, by Liouville's Theorem, this function $f(z)$ is $unbounded$.
Now we know that $|f(z)|$ $\to$ $\infty$, as $|z|$ $\to$ $\infty$.
But since we already established that $f$ is not a polynomial, then it does not have a pole at infinity. Then the singularity of $f$ at infinity must be an essential singularity.
Now we can apply Big Picard's theorem to claim that $f$ attains every complex number $a$, with the exception of at most one value, infinitely often.
Is this ok?
Thanks,
Using big Picard for this is major overkill.
If the image is not dense, there is a point $a$ and an $\varepsilon > 0$ such that $|f(z)-a| > \varepsilon$ for all $z$. Put $$ g(z) = \frac{1}{f(z)-a}. $$ Then $g$ is entire and bounded, thus constant. But this in turn forces $f$ to be constant, which is a contradiction.