Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an analytic, bounded function. Then $f$ is constant. ###### Proof Assume that $|f(z)| \leqslant M$ for all $z \in \mathbb{C}$. Let $a \neq b$ be two points in $\mathbb{C}$. Choose $R$ large enough so that $R >2 \max \{|a|,|b|\}$. ![[Drawing 2025-04-10 12.18.25.excalidraw|100]] That means that if we consider $w \in \partial B_R(0)$, that is $|w|=R$ then $|w-a|>\frac{R}{2}, \quad|w-b|>\frac{R}{2}$ Since $f$ is analytic in $\mathbb{C}$ we can use [[Cauchy's Integral Formula]] to compute $f(a)$ and $f(b)$ using $\partial B_R(0)$ as the curve $\gamma$ (of course positively oriented!). We have $ \begin{gathered} f(a)-f(b)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_R(0)} \frac{f(w)}{w-a} \mathrm{~d} w-\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_R(0)} \frac{f(w)}{w-b} \mathrm{~d} w \\ =\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_R(0)} f(w)\left(\frac{1}{w-a}-\frac{1}{w-b}\right) \mathrm{d} w=\frac{a-b}{2 \pi \mathrm{i}} \int_{\partial B_R(0)} \frac{f(w)}{(w-a)(w-b)} \mathrm{d} w . \end{gathered} $ Therefore $ |f(a)-f(b)| \leqslant \frac{|a-b|}{2 \pi} \frac{M}{R^2 / 4} \int_{\partial B_r(0)} 1|\mathrm{~d} w|=\frac{|a-b| 4 M}{R}, $ as $\int_{\partial B_R(0} 1|\mathrm{~d} w|$ is just the length of the curve, which equals $2 \pi R$. Notice that since $R$ is arbitrary (provided that it is big enough, as indicated above) we can send $R$ to infinity, showing that $|f(a)-f(b)|=0$ for any $a$ and $b$ in $\mathbb{C}$, therefore proving that the function is constant.