Rigidity of holomorphic function I'm trying to show that 
if $f=u+iv$ is continuous on the closed unit disk and holomorphic on the open unit disc and $u=v^2$ on the unit circle, then $f$ is constant.     
I was thinking to apply maximum and minimum principle for harmonic functions to $v^2-u$ but $v^2-u$ is not necessarily harmonic (it is subharmonic).
I also thought about  constructing a conformal map on the image of f that sends the parabola $x=y^2$ to the real line but I'm not sure if one exists.
 A: Let $U$ be the open unit disk.  The function $$g: \mathbb{C} \rightarrow \mathbb{R}$$ $$g(x + i y) = x - y^2$$ is open.  If $f$ is not constant then by the open mapping theorem $g \circ f(U)$ is open. The extreme values of $g \circ f$ on the compact set $\overline{U}$ are therefore attained on the unit circle.  Since the minimum is strictly less than the maximum $g \circ f$ is not constant on the unit circle.  A contradiction.
A: Let $D$ and $P$ as above in the solution of Micah.
Claim: $f(D)\subseteq P$.
Suppose there exists a $z_0\in D$ such that $f(z_0)\notin P$. Let's give a unit parametrization $\gamma(t)=e^{2\pi it}$ to unit circle. Then  $\sigma(t)=f(\gamma(t))$ is a closed curve contained in P. Note that $n(\sigma,f(z_0))=0$, where $n(,)$ denotes the index of a point with respect to a curve. Now
$n(\sigma,f(z_0))=\Sigma_{j=1}^{n}n(\gamma, z_j)$,
where $z_j$'s are the preimages of $f(z_0)$.
The RHS is a positive integer greater than $1$, and hence a contradiction. So the claim is true and hence $f$ has to be constant.
Now the only problem is that the statement 
$n(\sigma,f(z_0))=\Sigma_{j=1}^{n}n(\gamma, z_j)$ is true when $\gamma$ is a curve contained in a domain $\Omega$ and $f$ is holomorphic on $\Omega$. But as "it turns out for a function $f$ holo. in the interior of $D$ and cont. on D, this property is true."
Remark: Observe that there is nothing special in the Parabola here, one can choose any curve $\sigma$ such that $\mathbb{C}\setminus<\sigma>$ does not have any bounded component.
A: Let $D$ denote the closed unit disc; let $P=\{x+iy|x=y^2\}$. If $f(D) \subset P$, then $f(D)$ cannot be open. So $f$ is constant by the open mapping theorem.
On the other hand, suppose there is some $z_0 \in D$ with $f(z_0) \not\in P$. Notice that $P$ divides $\Bbb{C}$ into two non-compact path-components. On the other hand, $f(D)$ is compact (since $D$ is). So there must be some $w \not\in f(D)$ contained in the same path-component as $f(z_0)$. 
Take a path joining $f(z_0)$ to $w$ which does not intersect $P$. Then it must intersect the boundary of $f(D)$. Since $f(D)$ is compact, it is closed. So there is some $z_1$ such that $f(z_1)$ is contained in the boundary of $f(D)$ and not in $P$.
Since $f(z_1) \not\in P$, $z_1$ is not on the unit circle. Thus we must have $|z_1|<1$, so there is some open $U$ with $z_1 \in U \subset D$. By the open mapping theorem, if $f$ is non-constant, $f(U)$ is a $\Bbb{C}$-open subset of $f(D)$ containing $f(z_1)$. But $f(z_1)$ lies on the boundary of $f(D)$, so no such set exists. Thus $f$ must be constant.
