Let \(\Omega\subset\mathbb R^2\) be open and let \[ f:\Omega\to\mathbb R^2 \] be smooth. Assume that \(f\) preserves orientation and maps every pair of orthogonal curves to a pair of orthogonal curves.
Identifying \(\mathbb R^2\) with \(\mathbb C\), prove that \(f\) is holomorphic.
Solution
Write \(f=(u,v)\). Fix \(p\in\Omega\), and set \[ a = Df_p(1,0) = \bigl(u_x(p),v_x(p)\bigr), \qquad b = Df_p(0,1) = \bigl(u_y(p),v_y(p)\bigr). \]
1. The vectors \(a\) and \(b\) are orthogonal.
For \(\abs{t}\) small enough that both points lie in \(\Omega\), consider the curves \(t\mapsto p+t(1,0)\) and \(t\mapsto p+t(0,1)\). They pass through \(p\) with tangent vectors \((1,0)\) and \((0,1)\), so they are orthogonal at \(p\). Their images under \(f\) have tangent vectors \[ Df_p(1,0)=a \qquad\text{and}\qquad Df_p(0,1)=b. \] By the hypothesis that \(f\) maps orthogonal curves to orthogonal curves, \(a\cdot b=0\).
2. The vectors \(a\) and \(b\) have the same length.
The vectors \((1,1)\) and \((1,-1)\) are orthogonal. Curves through \(p\) with these tangent vectors have image tangent vectors \[ Df_p(1,1)=a+b, \qquad Df_p(1,-1)=a-b. \] Hence \[ 0 = (a+b)\cdot(a-b) = \norm{a}^2-\norm{b}^2. \] Therefore \(\norm{a}=\norm{b}\).
3. One has\[ b = \bigl(-v_x(p),u_x(p)\bigr). \]
Orientation preservation gives \[ \det Df_p = \det(a,b) > 0. \] In particular, \(a\neq0\). By steps 1 and 2, \(b\) is one of the two vectors obtained from \(a\) by a quarter-turn: \[ b = \pm\bigl(-v_x(p),u_x(p)\bigr). \] The positive choice satisfies \[ \det \begin{pmatrix} u_x(p)&-v_x(p)\\ v_x(p)&u_x(p) \end{pmatrix} = u_x(p)^2+v_x(p)^2 > 0, \] whereas the negative choice has negative determinant. Thus orientation preservation forces the positive choice.
4. The Cauchy–Riemann equations hold at \(p\):\[ u_x(p)=v_y(p), \qquad u_y(p)=-v_x(p). \]
By the definition of \(b\) and step 3, \[ \bigl(u_y(p),v_y(p)\bigr) = \bigl(-v_x(p),u_x(p)\bigr). \] Equality of the two coordinates gives the displayed equations.
5. The map \(f\) is\[ \boxed{\text{holomorphic on $\Omega$}}. \]
The point \(p\in\Omega\) was arbitrary, so step 4 gives the Cauchy–Riemann equations throughout \(\Omega\). Since \(f\) is smooth, the first partial derivatives of \(u\) and \(v\) are continuous. Therefore the Cauchy–Riemann criterion implies that the complex-valued map \(f=u+iv\) is holomorphic on \(\Omega\).
6. Q.E.D.
Step 5 is the required conclusion.