the cardinality of $S?$

Question

Let $S=\{ h:D \rightarrow D : h$ is analytic in $ D$ such that $h(z)^2=\overline {h(z)}$ for all $z\in D$}, Where $D=\{z \in \mathbb{C: \vert z \vert}<1\}$. Then the cardinality of $S?$

My work: let $f(z)=u+iv$ and using the given condition I find two relation

$$u^2-v^2=u$$

$$2uv+v=0$$ If suppose $v \neq0$ then $u=-1/2$ so $f$ is constant.Is the logic correct?

Answer 1

If $h$ is analytic, then so is $\,h^2=\overline h,\,$ and hence $\,h+\overline h,\,$ which is real valued, is also analytic. But, real valued and analytic, implies constant. Hence, $h$ is also constant.

So $h(z) \equiv w,\,$ with $\,w^2=\overline{w}.\,$ Thus $$\,|w|^2=|\overline w|=|w|,$$

and therefore, $\,|w|=0\,$ or $\,|w|=1.\,$

But $\,|w|=1\,$ is rejected, since it does not belong to $D$.

Hence $\,h\equiv 0\,$ and $\,|S|=1$.

Answer 2

Your approach has at least one problem. What do you mean by $v\neq0$? It happens that $v$ is a function, not a number. Do you mean that $v$ has no zores? Or that it is not the null function.

You can solve the problem using the fact that the only analytic functions $f$ such that $\overline f$ is also analytic are the constant function (I'm using here the fact that $D$ is connected). So, since $h^2$ is analytic, $\overline h$ is analytic and therefore $h$ is constante. So, this reduces the problem to the problem of finding all $w\in\mathbb D$ such that $w^2=\overline w$.