Let $\mathbb{D}$ denote the open unit disk. Fix $\lambda \in \mathbb{D}$. Define the Mobius transform $\phi_{\lambda}:\mathbb{D}\rightarrow\mathbb{D}$ by $$\phi_{\lambda}(z) = \frac{z-\lambda}{1-\overline{\lambda}z}$$ If possible, I'm trying to find the inverse of $\phi_{\lambda}^2$; that is, I want to find an analytic function $\psi:\mathbb{D}\rightarrow\mathbb{D}$ such that $$\phi_{\lambda}^2\Big(\psi(z)\Big) = \psi\Big(\phi_{\lambda}^2(z)\Big) = z$$

I attempted to find $\psi$ analytically by trying to solve the equation $$\phi_{\lambda}^2\Big(\psi(z)\Big) = z$$ but I know that "taking the square root on both sides" is not always a valid move in $\mathbb{C}$.

Any help will be appreciated.

  • $\begingroup$ I am afraid such an inverse might do not exist. Note that $f(z)=z^2:\mathbb{D}\to\mathbb{D}$ is onto but not one-to-one. Hence, $\phi_{\lambda}^2:\mathbb{D}\to\mathbb{D}$ is also onto but not one-to-one. $\endgroup$ – hypernova Jan 22 at 14:48

The Moebius transforms $$ f(z) = \frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\,.$$

Applied for your case, the Moebius transform $\phi_\lambda$ is associated to the matrix $$A _ \lambda =\begin{pmatrix} 1 & -\lambda \\ -\bar\lambda & 1 \end{pmatrix}. $$

You are searching for the inverse of $\phi_\lambda \circ \phi_\lambda$ which is the Moebius transform associated with the matrix $$ [(A_\lambda)^2 ]^{-1} = [(A_\lambda)^{-1}]^2 =\frac{1}{(1-|\lambda|^2)^2} \begin{pmatrix} 1 & \lambda \\ \bar\lambda & 1 \end{pmatrix}^2 =\frac{1}{(1-|\lambda|^2)^2}\begin{pmatrix} 1+ |\lambda|^2 &2 \lambda\\ 2 \bar\lambda& 1+ |\lambda|^2 \end{pmatrix} .$$ In other words, the inverse transform is given by $$\psi(z) = \frac{(1+ |\lambda|^2) z + 2 \lambda}{2\bar\lambda z + (1+ |\lambda|^2)}\,.$$


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