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I'm trying to prove the following result in complex analysis:

If $f $ is an analytic and bijective function from the unit disc to an open connected region $A$ then the distance from $f(0)$ to the the border of $A$ is at least $|f'(0)|$.

The theorems I think I have to use are Schwarz Lemma and the Maximum Modulus Principle, but I can't find a good way to put it all together for this problem.

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  • $\begingroup$ Are you sure that it shouldn't be "at most $|f'(0)|$"? – $f(z) = (1-z)/(1+z)$ maps the unit disk conformally to the right halfplane, and $f'(0) = 2$. The distance of $f(0)$ to the border is $1$. $\endgroup$ – Martin R May 1 '16 at 2:10
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Unless I am mistaken, the distance from $f(0)$ to the border of $A$ is at most $|f'(0)|$:

Assume that $B_d(f(0)) \subset A$. Then $$ h(z) = f^{-1}(f(0) + dz) $$ maps the unit disk $\Bbb D$ into itself with $h(0) = 0$, and $$ h'(z) = \frac{d}{f'(f^{-1}(f(0) + dz)} \, . $$ Applying the Schwarz Lemma to $h$ gives $$ 1 \ge |h'(0)| = \frac{d}{|f'(0)|} $$ and therefore $d \le |f'(0)|$.

A lower bound of the distance from $f(0)$ to the boundary of $A$ is $\frac 14 |f'(0)|$ according to the Koebe quarter theorem.

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  • $\begingroup$ how did you get the d into the derivative of h? $\endgroup$ – Mathem Kristel May 1 '16 at 4:07
  • $\begingroup$ @MathemKristel The chain rule. $d$ is the derivative of the inner function $z \to f(0)+dz $. $\endgroup$ – Martin R May 1 '16 at 4:41
  • $\begingroup$ oh, I get it! at first I thought dz was a partial derivative. Sorry, I got confused. $\endgroup$ – Mathem Kristel May 1 '16 at 5:19
  • $\begingroup$ @MathemKristel: No problem! $d$ is the radius of the disk $B_d(f(0))$, $d$ for "distance". $\endgroup$ – Martin R May 1 '16 at 6:01

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