If $f: \mathbb{C} \rightarrow \mathbb{C}$ is holomorphic and $|f(z)| \geq 1 \space \space \forall z \in \mathbb{C}$, show that $f$ is constant
I have attempted to prove this, but I am not sure if my proof is correct. I tried to follow the proof of Liouville's Theorem to do this:
Since $f$ is holomorphic, by Taylor's Thm, we have that $\forall z \in \mathbb{C}$ and $R>0$,
$$f'(z) = \frac{1}{2\pi i} \oint_{|w-z|=R} \frac{f(w)}{(w-z)^2} dw$$ Now using the estimation lemma, $$\left|f'(z) \right|\leq \frac{1}{2\pi} \oint_{|w-z|=R} \frac{|f(w)|}{R^2} |dw|$$ Then
$$\left|f'(z) \right|\leq -\frac{1}{2\pi R^2} \oint_{|w-z|=R} -\space|f(w)| |dw|$$ $$\leq-\frac{1}{2\pi R^2} \oint_{|w-z|=R} 1 |dw| = -\frac{1}{R}$$
Letting $R\rightarrow \infty$, we have $f'(z) = 0 \space \forall z \in \mathbb{C}$, i.e. $f$ is constant.
Is this correct?
Here is a simpler proof:
$|f(z)| \geq 1$ implies $f(z)\ne 0$ for all $z$. Therefore, $g(z)=\dfrac{1}{f(z)}$ is an entire function.
Since $|g(z)| \leq 1$ for all $z$, Liouville's Theorem says that $g$ is constant and so is $f$.