Existence (?) of an analytic $f$ such that $|f(z)|=e^{|z|},~|z|<1.$

Question

Prove or disprove: there is an analytic function $f$ on $D=\{z:~|z|<1\}$ such that $|f(z)|=e^{|z|},~z\in D.$

Attempt. I believe that such a function does not exist (besides, a candidate function would be for example $e^z$, but $|e^z|=e^{Rez}\leq e^{|z|}$, although I can not provide a proof of my claim.

Thanks in advance!

Answer 1

As $|f|\geq 1$ on $\mathbb{D}$, we can consider the analytic function $g=\dfrac{1}{f}$ on $\mathbb{D}$. Clearly $g \leq 1$ on $\mathbb{D}$, and $g(0)=1$. By Maximum Modulus Principle, $g$ and therefore $f$ is constant.

Answer 2

If such a $f$ exists, then $f$ non-zero, analytic $\implies \log f$ analytic $\implies \log |f| = \Re \log f$ harmonic.
However, $|z|$ isn't harmonic. This means $f$ doesn't exist.