Prove that there exists a non-constant analytic function $f$ such that $|f(z)|\le1$ on at least 2/3 of the complex plane

Question

I am taking complex analysis this year and the exam is coming up. We were provided with some previous exams and this question is in one of them.

It does seem extremely counter-intuitive, since if $f(z)$ is analytic and non-constant, it is unbounded on the whole of $\mathbb{C}$. If it is unbounded, then its absolute value (in my mind) must be larger than $1$ at infinitely many points, while its absolute value is less than $1$ at finitely many points. Of course, my logic fails somewhere since the question is asking to prove such a function exists. How would I go about doing so?

I tried taking $f(z) = 1/z$ and that works, except it is not analytic everywhere (namely, at $z=0$). I am at a loss. Functions with poles seem to be the only way out, since they're unbounded in a finite region of $\mathbb{C}$ and can be constrained to be bounded by any number everywhere else. Of course, not possible, since they're not analytic. How do I proceed? Thank you for the help.

Edit: Just thought of a possible way to achieve this, it involves having a function be bounded on, let's say, the first three quadrants of the complex plane and letting it run wild in the last one. This would make it obey the requirement. Of course, this seems very synthetic and I don't think it would make an analytic function anyways. Weird. One example that could work would be $f(z) = e^z$, the problem being that it is bounded on just $1/2$ of the plane, give or take...

Edit#2: The question also gives the definition of the 2/3 of the plane part:

Let's define a set $$K(a)=\{z:|Im(z)|\leq a/2, |Re(z)|\leq a/2, |f(z)|\leq 1\}$$ We need its area to be no less than $2a^2/3$ for sufficiently large $a$.

Answer 1

As you know, $\exp$ is entire and maps the left half-plane $H^{-}$ into the unit disk. Two hints in the hope of not depriving future readers of the fun of constructing an example:

1. What part $D$ of the plane does $\exp$ map into $H^{-}$?

2. Can we find an entire function mapping $H^{-} \cup D$ into $H^{-}$?

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Added: Here's the solution I found, with spoilers hidden. For definiteness, let's agree $H^{-}$ is the closed half plane.

• The set $D$ consists of:

all complex numbers $z = x + iy$ such that $\cos y \leq 0$. This is the infinite union of strips $\frac{\pi}{2} + 2k\pi \leq y \leq \frac{3\pi}{2} + 2k\pi$ with $k$ an integer, and comprises half the plane in the sense of the question. The union $H^{-} \cup D$ comprises three-quarters of the plane.

• An entire function mapping $H^{-} \cup D$ to $H^{-}$ is:

$z \mapsto \exp(z) - 1$: The exponential function $\exp$ maps $H^{-}$ to the unit disk and maps $D$ to $H^{-}$. Subtracting $1$ slides the image one unit to the left, yielding a subset of $H^{-}$.

• An entire function with the desired property is:

$f(z) = \exp(\exp(z) - 1)$.