Does the domain where a harmonic map is conformal has Hausdorff dimension smaller than one? Let $\mathbb{D}^2$ be the closed unit disk, and let $f:\mathbb{D}^2 \to \mathbb{R}^2$ be a harmonic map, satisfying $\det df \neq 0$ almost everywhere on $\mathbb{D}^2$. Suppose also that 
$$U= \{ p \in \mathbb{D}^2 \, | \, \det(df_p)<0  \, \, \text{and } df_p \text{ is conformal}\}$$
has measure zero in $\mathbb{D}^2$. (By "conformal" here I mean that the two singular values of $df$ are equal.)

Does $\dim_{\mathcal{H}}(U) < 1 $ ? ( $\dim_{\mathcal{H}}(U)$ is the Hausdorff dimension of $U$). 

An "easier" question: Is $U$ always finite?
I know that $\dim_{\mathcal{H}}(U) \le 1 $. The question is whether or not strict inequality holds.
($\dim_{\mathcal{H}}(U) \le 1 $ since the domain of conformality is the zero-set of a non-zero real analytic function).
 A: For some reason I had hard time parsing your question. First, some terminology: A point $p$ where $f$ has negative Jacobian determinant and is conformal in your sense should be called a point where $f$ is anticonformal as the above conditions are equivalent to
$$
\partial f(p)= 0, \bar\partial f(p)\ne 0. 
$$
(In other words, the function $\bar{f}$ is conformal at $p$ in the traditional sense.) Here
$$
\partial= \frac{\partial}{\partial z}, \quad \bar\partial= \frac{\partial}{\partial \bar{z}}. 
$$
The harmonicity condition for $f$ says 
$$
\bar\partial \partial f=0, \hbox{in} ~D^2. 
$$
Now, suppose that $p$ is a point at which $f$ has strictly negative Jacobian. This implies that
$$
h= \partial f
$$
satisfies $\bar\partial h=0$, i.e. $h$ is holomorphic in a (connected) neighborhood $V$ of $p$. Thus, this function $h$ is either constant in $V$ or has only finitely many zeroes in $V$. In other words, either $V\subset U$ or $V\cap U$ is finite. Since you are assuming that $U$ has empty interior, it follows that $V\cap U$ is finite, i.e.  $f$ is anticonformal in a finite subset of $V$. Thus, in view of compactness of $D^2$, your subset $U$ is finite. 
