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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 $z\in \mathbb{C}$?

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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 $c>1$.

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No, there are no such functions with $c>1$. Your condition means that they expand the spherical metric. If you look at the area of the whole sphere, you obtain a contradiction.

More precisely, looking at the area of the sphere and the area of its image we conclude that the area of the image is finite, therefore your function is rational (and $c$ is its degree). But a rational function of degree $>1$ must have a critical point the derivative is 0 at this point, which is in contradiction with your assumption.

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    $\begingroup$ If you take $w=z^2$, you get formula $$\int_{\mathbb{C}}|w'(z)|^2/(1+|w(z)|^2)^2 d x dy = 2\pi > \int_{\mathbb{C}}1/(1+|z|^2)^2 d x dy=\pi$$, so why this is a contradiction? $\endgroup$
    – Lira
    Commented Feb 21, 2020 at 6:57
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    $\begingroup$ Because $(dw/dz)(0)=0$ in your example, while the assumption excludes this. $\endgroup$ Commented Feb 21, 2020 at 12:53

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