Suppose that $f : \mathbb{C} \rightarrow \mathbb{C}$ is an entire function such that $|f(z)| \leq 1 + \sqrt{|z|}$. Show that $f$ is a constant function.
To do this, I showed that $|f'(z_{0})| = 0$, $\forall z_{0} \in \mathbb{C}$. Let $\gamma(t) = z_{0} + Re^{it} : t \in [0,2\pi] $.
Namely, Cauchy's Integral Formula says that $|f'(z_{0})| \leq \left| \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z-z_{0})^{2}} \mathrm{d}z \right| = \left| \frac{1}{2\pi} \int_{0}^{2\pi} \frac{f(z_{0} + Re^{it})}{Re^{it}} \mathrm{d}t \right| \leq \max_{t \in [0, 2\pi] } \left| \frac{f(z_{0} + Re^{it})}{Re^{it}} \right| \leq \max_{t \in [0, 2\pi] } \left| \frac{1 + \sqrt{|z_{0} + Re^{it}|}}{Re^{it}} \right|$
When $R > |z_{0}|$. Then $\max_{t \in [0, 2\pi] } \left| \frac{1 + \sqrt{|z_{0} + Re^{it}|}}{Re^{it}} \right| \leq \frac{1 + \sqrt{2R}}{R} \rightarrow 0$, as $R \rightarrow \infty$. I was wondering if this solution is correct.