Artificial intelligent assistant

complex analysis (Univalent function ) The Distortion Theorem tells us that if $f$ is a univalent function on $\mathbb{D}:=\\{z:|z|<1\\}$, then $|f'(z)|\leq 12\,|f'(0)|$ for $|z|\leq\frac12$. By iterating this, prove that if $f:\mathbb{H}\rightarrow D$ is any conformal transformation where $\mathbb{H}=\\{z:\Im(z)>0\\}$ and $\Im(z),\Im(w)\geq y>0$, then $$|f'(w)|\leq 144_{\vphantom{!}}^{\left(\tfrac{|z-w|}{y}\right)+1}\,|f'(z)|$$

Gah, what a horribly wasteful way to estimate the derivative. You are basically asked to go from $z$ to $w$ in horizontal steps. This is not how one should move about in the half-plane model of hyperbolic plane.

Oh well. Begin with the estimate $$|f'(\zeta)|\le 12 |f'(z)|,\qquad\text{where }\ |z-\zeta| \le \frac12 y\tag{1}$$ This follows from the fact that $\mathbb H$ contains the disk of radius $y$ centered at $z$. Inequality (1) is one step of the process.

Next, divide the line segment from $z$ to $w$ into sub-segments of length at most $y/2$. Say, you have $n$ sub-segments; then applying (1) $n$ times to the adjacent partition points, moving from $z$ to $w$, you get $$|f'(w)|\le 12^n |f'(z)|\tag{2}$$

It remains to estimate $n$. Since $ \left\lceil \frac{|z-w|}{y/2}\right \rceil $ segments are enough, the inequality $$n\le \frac{2|z-w|}{y}+1\tag{3}$$ holds. Done.

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