Question about Schwarz Lemma applied to comformal automorphism of a square Let $f$ be a conformal automorphism of the square $\{x+iy \mid x, y \in [-3,3] \} $ and suppose $f(0) = 2+2i$, show $\frac{1}{9\sqrt 2}  < |f’(0)| < \frac{10}{9\sqrt 2}$
I’ve tried to consider $f^*$ as a conformal automorphism of a disk containing the square, that coincides with $f$, and got an upper bound of $\frac{5}{9}$. However, I am not sure if such thing exist, and if it does, how to proceed to get the lower bound. 
 A: $f$ is not necessarily the restriction of an automorphism of a disk containing the square. But you can proceed as follows: Since
$$
 B(0, 3) \subset \{x+iy \mid x, y \in [-3,3] \} \subset B(0, 3\sqrt2)
$$
we can define
$$ 
g: \Bbb D \to \Bbb D \,,\quad  g(z) = \frac{f(3z)}{3 \sqrt 2}
$$
where $\Bbb D$ is the unit disk. Then $g(0) = \frac{2+2i}{3 \sqrt 2}$ and the Schwarz-Pick lemma gives
$$ 
 \frac{|f'(0)|}{ \sqrt 2} = |g'(0)| \le 1 - |g(0)|^2 = 1 - \left| \frac{2+2i}{3 \sqrt 2}\right|^2 = \frac 59 \\
\implies |f'(0)| \le \frac{5 \sqrt 2}{9} \, .
$$

For the lower bound we use that
$$
 (f^{-1})' (z) = \frac {1}{f'(f^{-1}(z)}
$$
and in particular
$$
 (f^{-1})'(2+2i) = \frac{1}{f'(0)}
$$
so that a lower bound for $|f'(0)|$ is equivalent to an upper bound for $|(f^{-1})'(2+2i)|$. And that can be obtained similarly as above: Since
$$
 B(2+2i, 1) \subset \{x+iy \mid x, y \in [-3,3] \} \subset B(0, 3\sqrt2)
$$
we can define
$$
h: \Bbb D \to \Bbb D \,,\quad  h(z) = \frac{f^{-1}(z+2+2i)}{3 \sqrt 2}
$$
Then $h(0) = 0$ and the Schwarz lemma gives
$$
 1 \ge |h'(0)| =  \frac{|(f^{-1})'(2+2i)|}{3 \sqrt 2} = \frac{1}{3 \sqrt 2 |f'(0)|} \\
\implies
 |f'(0)|  \ge \frac{1}{3 \sqrt 2} \, .
$$
