Is there is a holomorphic function $g:\mathbb{C}\to \mathbb{C}$ so that $$\frac{|g'(z)|}{1+|g(z)|^2}=\frac{c}{1+|z|^2}$$ for some $c>1$ and all $z\in \mathbb{C}$?



Analytic function on $\mathbb{C}$
There is no such function. Indeed, $h(z):=g(1/z)$ satisfies the same functional equation
$$\frac{|h'(z)|}{1+|h(z)|^2}=\frac{c}{1+|z|^2}.$$
In particular, $|h'(z)|>c/2$ for $|z|<1$, hence $h'(z)$ does not have an essential singularity at $z=0$. So $g(z)$ is a polynomial, and then letting $z\to\infty$ in the original functional equation shows that $g(z)=az+b$ with $|a|=1/c$. Then, in the original functional equation, the maximum of the left hand side equals $1/c$, while the maximum of the right hand side equals $c$. So $1/c=c$, which contradicts $c>1$.


