Holomorphic mapping on open unit disk

Assume that $$g$$ is a holomorphic function on a neighborhood of $$\overline{B}(0,1)$$ (the closed unit disc centered at $$0$$) such that $$g(0)=0$$. Let $$s=sup_{|z|=1}Re(g(z))$$. Consider the function $$h(z)=\frac{g(z)}{2s-g(z)}$$.
Show that the function $$h$$ maps $${B}(0,1)$$ (the open unit disk centered at $$0$$) to $${B}(0,1)$$ and that $$h(0)=0$$.

Any hint will be greatly appreciated. Thanks!

Assuming $$g$$ not identically zero (as otherwise $$h$$ doesn't make sense), we have by maximum modulus that $$\Re g(z) < s, |z|<1$$ and of course $$s>0$$ since $$s > \Re g(0)=0$$
But then $$|2s-g(z)|^2-|g(z)|^2=4s^2-4s\Re g(z)=4s(s-\Re g(z)) >0$$ for all $$|z|<1$$ so $$|g(z)| <|2s-g(z)|$$ hence $$|h(z)|<1, |z|<1$$, while $$h(0)=0$$ by defintion so we are done!
• Great, thank you! If I may ask, does this imply that the function $h$ is holomorphic?
• yes as it is a ratio of holomorphic functions (denominator nonvanishing anywhere in $|z|<1$) Commented Dec 14, 2020 at 23:32