Let \(f(z) = u(x,y) + i v(x,y)\) be analytic in a domain \(D \subset \mathbb{C}\). Let \(z_0 = (x_0, y_0) \in D\) be a point in the intersection of the level curves \(u(x,y) = c_1\) and \(v(x,y) = c_2\). Suppose that \(f'(z_0) \ne 0\). Prove that the lines tangent to these curves at \(z_0\) are perpendicular.
Let \(f(z) = z^2\) on \(\mathbb{C}\).
- Describe the level curves of \(\operatorname{Re}(f)\) and of \(\operatorname{Im}(f)\).
- What are the angles of intersection between the level curves \(\operatorname{Re}(f) = 0\) and \(\operatorname{Im}(f) = 0\)? Is your answer in agreement with part (a)?
Solution
Write \[ f=u+iv. \] At a point where \(f'(z_0)\ne0\), neither \(\nabla u(z_0)\) nor \(\nabla v(z_0)\) vanishes. By the Cauchy–Riemann equations, \[ \nabla u=(u_x,u_y), \qquad \nabla v=(v_x,v_y)=(-u_y,u_x). \] Therefore \[ \nabla u\cdot\nabla v =u_x(-u_y)+u_yu_x=0. \] The gradients are normal to the corresponding regular level curves, so their tangent lines are perpendicular.
For \(f(z)=z^2\), \[ u(x,y)=x^2-y^2, \qquad v(x,y)=2xy. \] Thus \(u=c\) gives rectangular hyperbolas \(x^2-y^2=c\); when \(c=0\) this is the pair of lines \(y=\pm x\). Likewise \(v=c\) gives \(2xy=c\); when \(c=0\) this is the pair of coordinate axes.
At the origin, the zero level sets therefore meet in lines separated by angles \(\pi/4\). This does not contradict part (a), because \[ f'(0)=0, \] so the level sets are not regular there and the hypothesis of part (a) fails. At every noncritical intersection, the two level curves meet orthogonally.