Let $$A = \begin{pmatrix} \cos\theta & b \\ -b & c \end{pmatrix} \in M_2 $$ be a contraction, i.e., $\Vert A\Vert\le 1$, and $$\gamma := \frac{b}{\sin\theta}\leq 1$$ Show that there exists $a \in\mathbb{C}$ with $\vert a\vert\leq 1$ such that $c=\gamma^2\cos\theta+(1-\gamma^2)a$.
It seems there always exists a point $a$ s.t. $c$can be written as a particular convex combination of $\cos\theta$ and $a$. But now the question is whether this $a$ will be in the unit disc? For that, I gave the following attempts.
Attempt I: I was initially trying to show $a\in W(A)$ i.e. numerical range of $A$ which would prove the required thing as $A$ is a contraction. But later I observed that $a$ could not belong to $W(A)$.
Attempt II: I tried to find out two vectors $x$ and $y$ with $\Vert x\Vert\le 1$ and $\Vert y \Vert \le 1$ s.t. $\langle Ax, y\rangle=a$ which would prove the required thing as $A$ is a contraction. Here also I observed later that it is not possible.
Attempt III: I also tried to show it by contrapositive argument i.e. if $\vert a \vert\ >1$ then there is a vector $x$ with $\Vert x\Vert\le 1$ s.t. $\Vert Ax\Vert >1$ but unfortunately could not get success here also.
Any hint/comment is highly appreciated. Thanks in advance.