Is this a valid proof of the orthogonality of harmonic conjugates?

My textbook (Churchill) is asking me to prove that the contours $u(x,y) = c_1$ and $v(x, y) = c_2$, where $u$ and $v$ are the real and imaginary components of an analytic function $f(z)$, are orthogonal at any point by noting that $u_x + u_y \frac{dy}{dx} = 0$ and then showing that the tangent lines are orthogonal.

I can do this easily via taking the dot product of $\nabla u$ and $\nabla v$ and using the Cauchy-Riemann equations to show that $\nabla u·\nabla v = 0$, hence the normal vectors of the contours are orthogonal, so the contours themselves are as well. But they asked to find the tangent line first, so here's what I tried, I haven't done any vector calculus in a few years though so I'm not sure if it's valid:

$$u_x + u_y \frac{dy}{dx} = \langle u_x,u_y \rangle · \langle 1, \tfrac{dy}{dx} \rangle = 0$$

$\langle u_x, u_y \rangle$ is $\nabla u$ which is normal to the contour, so $\langle 1, \frac{dy}{dx} \rangle$ must be tangent to the $u$ contour. From the above, $\frac{dy}{dx} = -\frac{u_x}{u_y}$, so $\langle 1, -\frac{u_x}{u_y} \rangle$ is tangent to the $u$ contour.

Similarly, $$v_x + v_y \frac{dy}{dx} = \langle v_x, v_y \rangle · \langle 1, \tfrac{dy}{dx} \rangle = 0$$

So $\langle 1, \frac{dy}{dx} \rangle$ is tangent to the $v$ contour. In this case $\frac{dy}{dx} = -\frac{v_x}{v_y}$, so the tangent vector is $\langle 1,-\frac{v_x}{v_y} \rangle$ which is equal to $\langle 1,\frac{u_y}{u_x} \rangle$ due to the Cauchy-Riemann equations. Then $\langle 1, -\frac{u_x}{u_y} \rangle · \langle 1, \frac{u_y}{u_x} \rangle = 1-1 = 0$, so the $u$ and $v$ contours are orthogonal.

Is this reasoning valid?

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If you use \langle x, y \rangle you'll produce $\langle x,y \rangle$. –  Pragabhava Dec 4 '12 at 3:09
Ah, much better, thanks! –  user46080 Dec 4 '12 at 3:13