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Let $Q$ be the unit square and $f:B(0,1) \rightarrow Q$ biholomorphic with $f(0)=0$. Show that $f(iz)=if(z)$ $ \forall z\in B(0,1)$ and follow that $f$ has a power series of the form $\sum_{n=0}^{\infty} a_n z^{4n+1}$

I tried the following:

I proved that a biholomorphic map is also conformal. Meaning it conserves angles.

Therefore, \begin{align*} \angle(iz, z) = 90° \Rightarrow \angle(f(iz),f(z)) \overset{!}{=}90° \end{align*}

In my opinion, $f(iz)=if(z)$ solves that condition, but so could any $u = \lambda f(z) \in \mathbb{C}$ with $\lambda \in \mathbb{R}$ since conformal maps conserve angles, but not lengths. So I am probably missing something here...

Next, I tried to calculate the Taylor series at $z_0 = iz$. I calculated the first few derivatives and $(z-iz)$... \begin{align*} &f(z) = -if(iz) \quad \text{and}\quad (z-iz)^0 = 1\\ &f'(z) = f'(iz) \quad \text{and}\quad (z-iz)^1 = (1-i)z\\ &f''(z) = if''(iz) \quad \text{and}\quad (z-iz)^2 = -2iz^2\\ &f'''(z) = -f'''(iz) \quad \text{and}\quad (z-iz)^3 = -2(i+1)z^3 \end{align*} Thus,

\begin{align*} f(z) = \sum_{n=0}^{\infty} \frac{f^n(iz)}{n!} (z-iz)^n = -if(iz) + f'(iz)(1-i)z+2f''(iz)z^2+2f'''(iz)(i+1)z^3 \end{align*}

Here I'm currently stuck. I tried to get information about the derivatives but I had no success, so I'm not even sure if I'm using the right approach. Can someone help me here?

Quotenbanane
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  • I assume that $Q$ is a unit square centered at the origin? Otherwise $f(0) = 0$ makes no sense. – Martin R Jan 16 '23 at 15:57
  • @MartinR yes exactly. – Quotenbanane Jan 16 '23 at 15:58
  • @MartinR One could define it as $Q = {z\in \mathbb{C} : |\text{Re}(z)|<1, |\text{Im}(z)|<1}$.

    I just realized the unit square is something else. Sorry for that.

     – Quotenbanane Jan 16 '23 at 15:59
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    This answers your question: analytic functions from square to unit disk because it proves that $f^{-1}$ satisfies your equation, hence $f$ also does. The subsidiary question then becomes obvious by identifying the coefficients of the two power series $f(iz)$ and $if(z).$ – Anne Bauval Jan 16 '23 at 16:12
  • @AnneBauval: The question you are linking to is about the mapping in the other direction (from the square to the unit disk). Of course it is closely related, so I will not object if the question is closed. – Martin R Jan 16 '23 at 16:16
  • @MartinR I know, whence my explanation about how it is related. It is so close that I think it may really be considered as a duplicate. – Anne Bauval Jan 16 '23 at 16:17
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    @AnneBauval: It seems to me that the second part is not so obvious to OP. But as I said, I will accept other decisions. – Martin R Jan 16 '23 at 16:22
  • For Quotenbanane, what I called obvious: if $\forall z\in B(0,1)\quad\sum_{n\in\Bbb N}a_ni^nz^n=\sum_{n\in\Bbb N}a_niz^n$ then $\forall n\in\Bbb N\quad a_ni^n=a_ni,$ i.e. $a_n=0$ when $n\not\equiv1\bmod4.$ – Anne Bauval Jan 16 '23 at 16:28
  • @MartinR Thank you both. At that time it definitely wasn't obvious to me but now it is! – Quotenbanane Jan 17 '23 at 11:42

1 Answers1

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The first part uses the symmetry of the domain $Q$: Show that $$ \phi(z) = f^{-1}\left( \frac 1i f(iz)\right) $$ is holomorphic in the unit disk, with $\phi(0) = 0$ and $\phi'(0) = 1$. Conclude that $\phi$ is the identity (hint: Schwarz lemma) and therefore $$ f(iz) = i f(z) $$ for all $z \in B(0, 1)$. Then substitute this identity in the Taylor series at $z_0 = 0$: $$ f(z) = \sum_{k=0}^\infty b_k z^k $$ and conclude all coefficients $b_k$ are zero, except if $k = 4n+1$ for some integer $n$.

Anne Bauval
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Martin R
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