Let $f$ be an entire function such that $f(0)=\dfrac{1}{2}$ and $|f(z)|\leq |e^z-\dfrac{1}{2}|$ for all $z \in \mathbb C$. Prove that $f(z)=e^z-\dfrac{1}{2}$ for all $z \in \mathbb C$.
Suppose $f \neq e^z-\dfrac{1}{2}$, note that since $f(0)=\dfrac{1}{2}$, then $f(z) \neq \dfrac{1}{2}-e^z$. It follows $|f(z)|<|e^z-\dfrac{1}{2}|$. Since we are under the hypothesis of Rouché's theorem, we can affirm that $e^z-\dfrac{1}{2}$ and $e^z-\dfrac{1}{2}+f(z)$ have the same number of zeros in all $\mathbb C$.
I am trying to arrive to an absurd but I can't, I would appreciate if someone could help me to arrive to the conclusion $f(z)=e^z-\dfrac{1}{2}$.
