# Prove that These Families of Level Curves are Orthogonal

From p. 79 in Brown's and Churchill's "Complex Variable and Application":

Let the function $f(z) = u(x, y)+iv(x, y)$ be analytic in a domain $D$, and consider the family of level curves $u(x, y) = c_1$ and $v(x, y) = c_2$. Prove that these families are orthogonal.
Specifically, show that if $z_0 = (x_0, y_0)$ is a point in $D$ which is common to two particular curves $u(x, y) = c_1$ and $v(x, y) = c_2$, and if $f'(z_0) \ne 0$, then the lines tangent to the curves at $(x_0, y_0)$ are perpendicular to each other. Then, the question gave a clue that:

$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx} = 0$ and $\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}\frac{dy}{dx} = 0$ (*)

At first, I thought that the authors meant $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}\frac{dx}{dx}+\frac{\partial u}{\partial y}\frac{dy}{dx} = 0$, $\frac{\partial v}{\partial x} = \frac{\partial v}{\partial x}\frac{dx}{dx}+\frac{\partial v}{\partial y}\frac{dy}{dx} = 0$. However, that cannot be the case, since we were told that $f'(z_0) = u_x(x_0, y_0)+iv_x(x_0, y_0) \ne 0$. Furthermore, $\frac{\partial u}{\partial x} = 0$ implies that $u(x, y)$ is constant in a direction parallel to the $x$-axis, but that certainly is not the case in general. So, how do you get the two equalities in (*)?

Furthermore, what is the significance in $f'(z_0) \ne 0$? In the next question, we are asked to show that with $f(z) = z^2$, the level curves $u(x, y) = x^2-y^2=0$ and $v(x,y)= 2xy = 0$ are not orthogonal. But a straight-forward computation $\nabla u · \nabla v = u_xv_x + u_yv_y = 4xy - 4xy = 0$, a constant zero. What did I gloss over?

Hint. To do this problem efficiently it is probably best if you take the point of view of two-variable real calculus. The level curves $u=c_1$ and $u=c_2$ are orthogonal at a certain point if their normal vectors are orthogonal at that point. These normals are $$\nabla u =\Bigl(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\Bigr) \quad\hbox{and}\quad \nabla v =\Bigl(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\Bigr)\ ,$$ provided they are not zero vectors. Can you finish the problem by explaining why these vectors are perpendicular in the context of your question?
• Actually, I figured out why the condition that $f'(z) \ne 0$ is necessary. If not, then $\nabla u, \nabla v$ are zero vectors, which have no directions. Nonetheless, what are the two equalities in (*)? $\nabla u$ is supposed to be normal to the line tangent to $u$ at $(x_0, y_0)$, but I am not seeing the expression in (*) as a directional vector. It looks more than $u_x$ to me. – Andy Tam Jun 30 '14 at 14:22
• If you regard $u(x,y)=c_1$ as defining $y$ implicitly in terms of $x$, then the first expression in $(*)$ is $du/dx$ (NB not $\partial u/\partial x$). – David Jun 30 '14 at 14:42