Since \(\phi(x,y)=K_1\) is an implicit equation of a plane curve, the gradient vector grad \(\phi\text{,}\) evaluated at \((a,b)\text{,}\) is perpendicular to the curve at \((a,b)\text{.}\) This vector is given by
\begin{equation*}
\mathbf{N}_1=\phi_x(a,b) +i\phi_y(a,b)
\end{equation*}
Similarly, the vector N\(_2\) defined by
\begin{equation*}
\mathbf{N}_2=\psi_x(a,b) +i\psi_y(a,b)
\end{equation*}
is orthogonal to the curve \(\psi(x,y) =K_2\) at \((a,b)\text{.}\) Using the Cauchy-Riemann equations, \(\phi_x=\psi_y\) and \(\phi_y=-\psi_x\text{,}\) we have
\begin{align}
\mathbf{N}_1 \cdot \mathbf{N}_2 \amp = \phi_x(a,b)\psi_x(a,b)+\phi_y(a,b)\psi_y(a,b)\tag{10.4.3}\\
\amp = -\phi_x (a,b) \phi_y(a,b) + \phi_y(a,b) \phi_x(a,b)\notag\\
\amp = 0\text{.}\notag
\end{align}
In addition, \(F,'(a+ib) \ne 0\text{,}\) so we have
\begin{equation*}
\phi_x(a,b) +i\psi_x(a,b) \ne 0\text{.}
\end{equation*}
The Cauchy-Riemann equations and the facts
\(\phi_x(a,b) \ne 0\) and
\(\psi_x(a,b) \ne 0\) imply that both
\(\mathbf{N}_1\) and
\(\mathbf{N}_2\) are nonzero. Therefore
Equation (10.4.3) implies that
\(\mathbf{N}_1\) is perpendicular to
\(\mathbf{N}_2\text{,}\) and hence the curves are orthogonal.