Is There Entire Function $|f(z)|=|1-|z||$ For All $z\in\mathbb{C}$

Question

Is There Entire Function $|f(z)|=|1-|z||$ For All $z\in\mathbb{C}$?

Looking at $z=e^{i\theta}$ we have $|f(z)|=|1-|e^{i\theta}||=0$

So this is an entire function with infinite zeros, which can not be by the identity theorem, So therefore there is not such entire function

Is this correct?

Answer 1

Not quite. Since $|z|=1\implies f(z)=0$, you can deduce from the identty theorem that $f$ is the null function. But this is impossible, since the null function doesn't satisfy the condition ftom the statement of the theorem.

Answer 2

Just for the sake of the exercise, I'd go with the maximum modulus principle and minimum modulus principle and assumption that such an entire function exists:

• For $|z|=1$, $\min|f(z)|$ and $\max|f(z)|$ is attained on $|z|=1$, thus $\color{red}{|f(z)|=0}, \forall z\in \overline{B(0,1)}$
• For $|z|=2$, $\min|f(z)|$ and $\max|f(z)|$ is attained on $|z|=2$, thus $\color{red}{|f(z)|=1}, \forall z\in \overline{B(0,2)}$

but $\overline{B(0,1)} \subset \overline{B(0,2)}$ and we have a contradiction, because $\forall z \in \overline{B(0,1)} \Rightarrow z \in \overline{B(0,2)}$ and $0=|f(z)|\ne |f(z)|=1$.