%The lens equation of equation  is in the vector formalism.  This may
%be simply transformed into the complex number formalism by introducing
%the complex numbers $z = x_1 + i x_2$ and $z_s = y_1 + i y_2$ 
%which are the positions in the deflecting and source planes respectively,
%and redefining $\vec{\alpha}(\vec{x}$) as a complex function.
% 
%The equivalent lens equation written with the complex number formalism is
%\begin{equation}
%\label{comple}
%z_s = z + \gamma \bar{z} - \frac{\epsilon}{\bar{z}},
%\end{equation}
%where $\epsilon = (\frac{\kappa_s}{1 - \kappa_c})$,
%$\kappa_s$ is the surface density of compact objects (ie. stars) and
%$\kappa_c$ is the surface density of continuously distributed matter.
%
%An expression for the
%Jacobian determinant, stated here without proof
%\begin{equation}
%\label{detAcomp}
%\det A = \left( \frac{\partial z_s}{\partial z} \right)^2 - 
%\left( \frac{\partial z_s}{\partial \bar{z}}\right) 
%{\left( \frac{\overline{\partial z_s}}{\partial z} \right) }.
%\end{equation}
%This may be used to find an expression for the critical curves, which 
%correspond to points where $\det A = 0$.
%
%Using equation (\ref{detAcomp}) in the complex lens equation (\ref{comple})
%gives
%\begin{equation}
%\det A = 1 - \left(\gamma + \frac{\epsilon}{\bar{z}^2}\right) \left(\gamma +
%\frac{\epsilon}{z^2} \right) = 0.
%\end{equation}
%This can be rewritten in polar coordinates, where $z =  x \cos \phi +
%i x  \sin \phi$, to get
%\begin{equation}
%x^4 (1-\gamma^2) - 2 \gamma \epsilon x^2(cos^2 \phi - sin^2 \phi) -1 = 0,
%\end{equation}
%which are Cassini ovals.
%This equation is  fourth order in $x$, but second order in $cos^2 \phi$,
%so a parametric form for the curves may be found by defining 
%\begin{equation}
%\lambda = cos^2 \phi - sin^2 \phi 
%\hspace{1cm} \mbox{and} \hspace{1cm} u = x^2.
%\end{equation}
%We now have a quadratic equation in $u$ 
%\begin{equation}
%u^2 (1 - \gamma^2) - 2\gamma\epsilon\lambda u - 1 = 0, 
%\end{equation}
%which has solutions
%\begin{equation}
%\label{usol}
%u = \frac{\gamma\epsilon\lambda \pm \sqrt{\gamma^2(\lambda^2-1) + 1}}
%	 {(1-\gamma^2)} .
%\label{u}
%\end{equation}
%In order for equation (\ref{usol}) to be a solution to the lens equation,
%the following conditions must be satisfied:
%\begin{itemize}
%\item $\gamma^2(\lambda^2-1) + 1 \geq 0$
%\item $\gamma^2 \neq 1$
%\item $u > 0$
%\end{itemize}
%For a given value of $\gamma$ then, there may be zero, one or two
%solutions for u and hence zero, two or four solutions for $x$.
%The caustics in the source plane, which correspond to these critical
%curves, are found by substituting $z = x e^{i \phi}$ into the 
%lens equation and using 
%$\cos \phi = \sqrt{\frac{1 + \lambda}{2}}$ and 
%$\sin \phi = \sqrt{\frac{1 - \lambda}{2}}$.
%Hence we have an expression for the caustics in terms of $z_s = y_1 + 
%i y_2$
%\begin{eqnarray}
%\label{cry1}
%y_1 & = & \left[ (1 + \gamma)x - \frac{\epsilon}{x} \right] 
%\sqrt{\frac{1 + \lambda}{2}}\\ 
%\label{cry2}
%y_2 & = & \left[ (1 - \gamma)x - \frac{\epsilon}{x} \right]
%\sqrt{\frac{1 - \lambda}{2}}
%\end{eqnarray}
%There are four representative regions within which solutions for the
%caustics may be found: $\gamma < -1$, $-1 <\gamma <0$, $0 < \gamma < 1$,
%$\gamma > 1$.    The lengths in the source plane
%are scaled by $\eta_0$ and in the deflecting plane by $\xi_0$.

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