If $f(z)^2+g(z)^2=1, \forall z \in A \subset R^2$ then $f_xg_y - f_yg_x=0, \forall z \in A$

Question

Let $A \subset R^2$ such that $B(1) \subset A$, $B(1)$ is the unit open ball.

Consider a $C^{\infty}$ function $h:A \to R^2$, $h(z)=(f(z),g(z))$ such that $f(z)^2+g(z)^2=1, \forall z \in A $

I need to prove that $f_xg_y - f_yg_x=0, \forall z \in A$

I tried to show that $det(Dh)=0$ by looking at $\omega=fdg-gdf$, since $d\omega=det(Dh)dx \wedge dy$. But couldn't manage to get any further. And I'm pretty much convinced this wouldn't lead me anywhere.

How can I use the fact that $f^2+g^2=1$?

Any advices?

Answer 1

Taking the partial derivatives of $f^2(z)+g^2(z)=1$ gives $$ f_x(z) f(z) + g_x(z) g(z) = 0 \\ f_y(z) f(z) + g_y(z) g(z) = 0 $$ For fixed $z \in A$ this is a homogeneous linear equation system for $(f(z), g(z))$ with the coefficient matrix $$ \begin{pmatrix} f_x(z) & g_x(z) \\ f_y(z) & g_y(z) \end{pmatrix} $$ which must have determinant zero because a non-trivial solution exists.