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?
 A: 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?
A: One sec. I think I realize how to use the hint (trying to answer my own question in comments).
If $u_x + u_y y' = 0$, then $v_y - v_x y'=0$ by Cauchy-Riemann. Here, we must have $v_x\neq 0$ for otherwise $v_y=0$ too, contradicting the nonzero assumption. Hence, we get $y'=v_y / v_x$ for the first level curve. Similarly, the second level curve gives $y'=-v_x/v_y$. Multiplying these gradients gives $-1$, proving orthogonality as desired.
Is this OK?
