Finding the inverse of $\phi_{\lambda}^2$ where $\phi_{\lambda}$ is the Mobius transform on $\mathbb{D}$

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.

• 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. – 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 $$\psi(z) = \frac{(1+ |\lambda|^2) z + 2 \lambda}{2\bar\lambda z + (1+ |\lambda|^2)}\,.$$