Proving an entire function with constant imaginary part on closed unit disc is constant on the whole $\mathbb C$ without identity principle

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.24

(Exer 8.24) If $f: \mathbb C \to \mathbb C$ is entire and $\Im(f)$ is constant on closed unit disc $\{|z| \le 1\}$, then $f$ is constant on $\mathbb C$.

In OSU course here, there's a proof of Exer 8.24 above where the proof makes use of the Identity Principle. I think I can prove alternatively by modifying the proof of Liouville's Thm instead of using the Identity Principle. I attempt my alternative proof as follows.

Question: What mistakes, if any, have I made, and why?

Pf:

By Cauchy-Riemann, $f \equiv: K$ is constant on closed unit disc. It remains to show $f \equiv: K$ on the rest of $\mathbb C$.

Consider any path $\gamma := \{ |z-w| = R \} \subset \mathbb C, w \in \mathbb C, R > 0$. We have by Cauchy's Integral Formula for $f'$ (Formula 5.1) that

$$|f'(w)| = |\frac{1}{2 \pi i} \int_{\gamma} \frac{f}{(z-w)^2} dz| = \frac{1}{2 \pi} |\int_{\gamma} \frac{f}{(z-w)^2} dz|$$

$$ \le \frac{1}{2 \pi} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \text{length}(\gamma) = \frac{1}{\not{2}\not{\pi}} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \not{2}\not{\pi} R$$

$$ = \max_{z \in \gamma}|\frac{f}{(z-w)^2}| R = \max_{z \in \gamma}\frac{|f|}{|z-w|^2} R = \max_{z \in \gamma}\frac{|f|}{R^2} R = \max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R} = \frac{K}{R}$$

$$\therefore, |f'(w)| = \lim_{R \to \infty} |f'(w)| \le \lim_{R \to \infty} \frac{K}{R} = 0 \ \forall w \in \mathbb C \implies f'(w) = 0 \ \forall w \in \mathbb C$$

$\therefore,$ by a theorem in the textbook (Thm 2.17), $f \equiv: K$, not only in closed unit disc, but also on the whole $\mathbb C$. QED

Answer 1

Here is another approach that uses the open mapping theorem.

Since $B(0,1)$ is open and $f$ is analytic, if $f$ is non constant, then $f(B(0,1))$ is open.

However, the set $\{x+iy| y = \operatorname{im} f(0) \}$ is not open, hence $f$ must be a constant.

Answer 2

The mistake is

$$\max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R}$$

We know that $f=|f|\equiv:K$ on closed unit disc, so including the unit circle. How do we know that $f=|f|\equiv:K$ on some $\gamma$ seemingly pulled out of a hat?

It seems that modifying Liouville's Thm will not work as it did (and should) for some exercises in Ch5. Instead, we prove by Identity Principle (Principle 8.15) as in the OSU homework:

Pf using Identity Principle (Principle 8.15):

After proving that $f$ is constant on closed unit disc, consider $g: \mathbb C \to \mathbb C$ s.t. $g(z) := f(z) - K$. Observe that $g$ is entire and zero on closed unit disc. Now consider a sequence of distinct complex numbers $\{a_n\}_{n=1}^{\infty}$ in the closed unit disc that converges to a complex number $a$ in the closed unit disc. Observe that $g(a_n)=0 \ \forall n \in \mathbb N$. $\because \mathbb C$ is a region, by Identity Principle (Principle 8.15), $g \equiv 0$ on $\mathbb C$. $\therefore, f(z) \equiv K$ on $\mathbb C$.

QED