Problem: Is there is an entire function $f$ such that $f(1)=\pi$ and $f'(z)=|z|f(z)$ for all $z\in\mathbb C$?
Question: Is the following correct?
On the unit circle line $\partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=\pi$ is uniquely solved by $f(z)=\frac{\pi}{e}e^z$. Since $\partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=\frac{\pi}{e}e^z$ on the entire $\mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $\mathbb C$. So, on $\mathbb{C}\setminus Z$, we would have $\lvert z\rvert=\frac{f'(z)}{f(z)}$. But $z\mapsto\lvert z\rvert$ is nowhere differentiable.