[proofplan] We prove that a bounded entire function is constant by showing $f' \equiv 0$. The [Cauchy integral formula](/theorems/345) for the first derivative, applied on circles of arbitrarily large radius $R$, combined with the ML inequality, gives $|f'(z)| \leq M/R$ for every $R > 0$. Sending $R \to \infty$ forces $f'(z) = 0$ at every point. Since $\mathbb{C}$ is connected, $f$ is constant. [/proofplan] [step:Bound $|f'(z)|$ using the Cauchy integral formula on a large circle] Fix $z \in \mathbb{C}$ and let $R > 0$. Since $f$ is entire, $\overline{B(z, R)} \subseteq \mathbb{C}$ and the [Cauchy integral formula](/theorems/345) for the first derivative gives \begin{align*} f'(z) = \frac{1}{2\pi i} \int_{\partial B(z, R)} \frac{f(w)}{(w - z)^2} \, dw. \end{align*} Apply the ML inequality with $\ell(\partial B(z, R)) = 2\pi R$ and $\sup_{w \in \partial B(z,R)} \left|\frac{f(w)}{(w-z)^2}\right| \leq \frac{M}{R^2}$ (using $|f(w)| \leq M$ and $|w - z| = R$): \begin{align*} |f'(z)| \leq \frac{1}{2\pi} \cdot 2\pi R \cdot \frac{M}{R^2} = \frac{M}{R}. \end{align*} [/step] [step:Send $R \to \infty$ to conclude $f' \equiv 0$ and $f$ is constant] Since $R > 0$ was arbitrary, letting $R \to \infty$ gives $|f'(z)| \leq \lim_{R \to \infty} M/R = 0$, so $f'(z) = 0$. This holds for every $z \in \mathbb{C}$, hence $f' \equiv 0$ on $\mathbb{C}$. Since $\mathbb{C}$ is a connected domain and $f' \equiv 0$, the function $f$ is constant. [guided] The bound $|f'(z)| \leq M/R$ is known as **Cauchy's estimate** for the first derivative. It is a direct consequence of the Cauchy integral formula and the ML inequality. The key point is that $R$ can be taken arbitrarily large because $f$ is entire (holomorphic on all of $\mathbb{C}$), so the formula applies on circles of any radius centred at any point. The conclusion $f' \equiv 0 \implies f$ constant uses the fact that $\mathbb{C}$ is connected. On a connected open set, a holomorphic function with identically zero derivative must be constant: for any two points $z_1, z_2 \in \mathbb{C}$, connect them by a path $\gamma$ and compute $f(z_2) - f(z_1) = \int_\gamma f'(z) \, dz = 0$. [/guided] [/step]