The equation of an ellipse in the complex plane is given by
$$ |z - f_1| + |z - f_2| = k $$
where $z$ is an arbitrary complex number, $f_1$ and $f_2$ are the foci, and $k$ is some constant such that $k \ge 2a$, $a$ being the length of the semi-major axis. This got me into thinking about what happens when $0 < k < 2a$.
While I haven't found an answer to that (though, I'd love to know), it struck me that if $k = |f_2 - f_1|$, it is supposed to represent the line segment joining $f_1$ and $f_2$.
My reasoning: $z$, $f_1$ and $f_2$ can be thought to form a triangle if $|z - f_1| + |z - f_2| > |f_2 - f_1|$. Now if the inequality is replaced by an equality, then $z$ must lie on the line joining $f_1$ and $f_2$. It must also be in between $f_1$ and $f_2$ because the equality won't hold otherwise.
Is this argument correct? And if it is, can I actually represent a line segment joining any two points $a$ and $b$ by this equation?
$$ |z - a| + |z - b| = |a - b|. $$
Also, can this equation be reduced to a form that can be handled algebraically?
You can suppose $b=0$ (otherwise, do a translation) and $a\ne0$. The equation is $$ |z|+|z-a|=|a| $$ Write $a=ru$, where $|u|=1$ and $r=|a|$, and $z=wu$. Then the equation is $$ |w|+|w-r|=r $$ Squaring gives $$ w\bar{w}+(w-r)(\bar{w}-r)+2|w|\,|w-r|=r^2 $$ and, simplifying, $$ 2|w|\,|w-r|=r(w+\bar{w})-2w\bar{w}\tag{*} $$ If we set $w=x+yi$, this requires $rx-(x^2+y^2)\ge0$. Under this condition, we can square (*): $$ 4w\bar{w}(w-r)(\bar{w}-r)= r^2(w^2+2w\bar{w}+\bar{w}^2)+4w^2\bar{w}^2-4rw\bar{w}(w+\bar{w}) $$ or, simplifying, $$ r^2(w-\bar{w})^2=0 $$ Therefore $w=\bar{w}$ is real and so $y=0$. The condition $rx-(x^2+y^2)\ge0$ now becomes $0\le x\le r$.
Thus $z=xu$, where $0\le x\le |a|$, so $z$ is a point on the segment connecting $0$ to $a$. Conversely, any such point satisfies the initial equation.