You can start with the parametrization
$$ x(s) = a \cos s $$
$$ y(s) = b \sin s $$
and write
$z(s) = x(s) + i y(s) = a (\cos s + i \sin s) + (b-a)(i \sin s)$
$ = a e^{i s} + \frac{b-a}{2}( e^{is}-e^{-is}) $
$ = (a + \frac{b-a}{2}) e^{i s} - (\frac{b-a}{2}) e^{-is}. $
$ = A_+ e^{is} + A_- e^{-is},$
with $A_\pm = \frac{a\pm b}{2} \in \mathbb{R}$.
There are several virtues to this representation. If you want the ellipse to be centered around a point $z_0 = x_0 + i y_0$, just add it:
$z(s) \rightarrow z_0 + A_+ e^{is} + A_- e^{-is}$;
If you want the ellipse to be rotated by an angle $\psi$, just multiply:
$z(s) \rightarrow e^{i \psi}(A_+ e^{is} + A_- e^{-is}) = A_+ e^{i(s+\psi)} + A_- e^{-i(s-\psi)}.$
Application
If you are an engineer like I am, you are probably thinking of these equations in terms of phasors, which are complex numbers with fixed magnitude and linear phase (their phase changes at a constant rate). We are in the business of turning real signals into complex phasors for DSP applications. To do this, one must build a filter that has the effect of adding a $\pi/2$ phase shift. If you are not careful about how you build this filter, you can end up with an additional (backwards propagating) phasor that is identical to the one above. The result is a signal that traces out an ellipse, not a circle, in the complex plane. If you goof up the phase shift and get it wrong by a small amount ($\pi/2-\epsilon$), this equivalent to the above parametrization with $$\frac{A_-}{A_+} = \tan (\epsilon/2).$$
(The ellipse will also be rotated by an angle $\psi = \pi/4$.)
Best Answer
The distance of the point $z$ from the line $b + ct,\; t \in \mathbb{R}$ is the length of the projection of $z-b$ to the normal, which has direction $\pm ic$. If we identify $\mathbb{C}$ with $\mathbb{R}^2$, we'd find the length of the projection by computing the inner product. We do the same in $\mathbb{C}$ even if we don't explicitly identify it with $\mathbb{R}^2$, the real inner product of $v$ and $w$, expressed in complex form, is $\Re \overline{v}w$.
So we get
$$\left\lvert\Re \left(\frac{\overline{ic}(z-b)}{\lvert c\rvert}\right)\right\rvert = \left\lvert\Im \frac{\overline{c}(z-b)}{\lvert c\rvert} \right\rvert$$
as the expression for the distance of $z$ from the line $b + ct$. If $c$ is chosen with absolute value $1$, that simplifies to $\lvert \Im \overline{c}(z-b)\rvert$.