\(f(z)= u(x,y) +i v(x,y)\) be analytic in a domain \(D\subset {\mathbb C}\). Let \(z_0=(x_0,y_0)\) be a point in \(D\) which is in the intersection of the curves \(u(x,y)= c_1\) and \(v(x,y)=c_2\), where \(c_1\) and \(c_2\) are constants. Suppose that \(f'(z_0)\neq 0\). Prove that the lines tangent to these curves at \(z_0\) are perpendicular.
Let \(f(z)=z^2\) be defined in \({\mathbb C}\).
Describe the level curves of \(\mbox{\textrm Re}{(f)}\) and of \(\mbox{Im}{(f)}\).
- What are the angles of intersections between the level curves \(\mbox{\textrm Re}{(f)}=0\) and \(\mbox{\textrm Im}{(f)}\)? Is your answer in agreement with part a) of this question?
Solution
Since \(f'(z_0)\ne0\), the Cauchy–Riemann equations give \[ \nabla u=(u_x,u_y), \qquad \nabla v=(v_x,v_y)=(-u_y,u_x), \] at \(z_0\). Thus \[ \nabla u\cdot\nabla v=-u_xu_y+u_yu_x=0, \] and both gradients are nonzero because \(|f'(z_0)|^2=u_x^2+u_y^2\). The tangent line to a regular level curve is orthogonal to its gradient, so the tangent lines to \(u=c_1\) and \(v=c_2\) are perpendicular.
For \(f(z)=z^2\), \[ u(x,y)=x^2-y^2, \qquad v(x,y)=2xy. \] Hence \(u=c\) consists of the rectangular hyperbolas \(x^2-y^2=c\), while \(v=c\) consists of the rectangular hyperbolas \(2xy=c\); the zero levels are the lines \(y=\pm x\) and the coordinate axes, respectively.
Part (ii) is incomplete in the source because it writes only ``\(\operatorname{Im}(f)\)’’ rather than specifying a level. Under the natural reading \(\operatorname{Im}(f)=0\), the curves \(\operatorname{Re}(f)=0\) are \(y=\pm x\), while \(\operatorname{Im}(f)=0\) gives the \(x\)- and \(y\)-axes. Every intersection angle is therefore \(\pi/4\) (or its supplementary angle \(3\pi/4\)). The origin is exceptional for part (a), since \(f'(0)=0\), so there is no contradiction with the orthogonality theorem there.