= Solution
The <Cauchy-Riemann equations> are $\boxed{u_x=v_y,\quad u_y=-v_x}$. They give the <gradient> identity
$$
\nabla u\cdot\nabla v=u_xv_x+u_yv_y=v_yv_x-v_xv_y=0.
$$
At a regular intersection, where $f'(z)\ne0$, the two nonzero <gradients> are <normal vectors> to the two <level sets>, so their tangent directions are perpendicular. This is the <orthogonality of regular real and imaginary level curves>.
The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f'|$, so at a <critical point> both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }f'(z)\ne0}$, rather than an unrestricted claim at critical points.
Back to article page
