= Solution
A <complex derivative> computed along the real and imaginary directions gives $f'=u_x+iv_x=v_y-iu_y$. Equality yields the <Cauchy-Riemann equations>
$$
\boxed{u_x=v_y,\qquad u_y=-v_x.}
$$
The <gradient>s satisfy $\nabla u\cdot\nabla v=u_xv_x+u_yv_y=0$. Since a <gradient> is normal to a regular <level set>, the two families meet at right angles by <orthogonality> wherever $f'\ne0$. At a <critical point> the regular-curve argument fails: for $f=z^2$ both <gradient>s vanish at zero, so the assertion requires this qualification.
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