Find $z$ such that $|z-w| - |z-w^2|$ is maximal where $z$ lies on the curve $|z-4| + |z+4| = 10$, where $w$ is a non real cube root of unity.
I found the locus of the $z$ from the given equation, which is an ellipse but am unable to proceed after it.
The locus that I found was $ \frac{x^2}{25} + \frac{y^2}{9} = 1$ where I assumed $z = x + iy$ .
I am a class 12 student who came across this problem in one of the online exams that I was supposed to give for my coaching.
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Denote $F_1, F_2$ the points with complex coordinates $F_1(-4),F_2(4)$ or, if you prefer, $F_1(-4+0i),F_2(4+0i).$ Then $$|z-4|+|z+4|=10$$ is equation of an ellipse with foci $F_1, F_2$ and $2a=10.$
Cubic roots of unity are $1,w,w^2,$ where $w=e^{i2\pi/3}=-\frac{1}{2}+i\frac{\sqrt3}{2}.$
For a fixed real $m$ satisfying $\;0<|2m|<|w-w^2|,$ $$|z-w| - |z-w^2|=2m \tag{H}$$ is equation of one branch of hyperbola with foci in $w$ and $w^2.$
We want to maximize $|z-w| - |z-w^2|,$ thus we assume that $m>0$ and we maximize $m.$ The corresponding branches of hyperbolas are lower.
The common center of hyperbolas lies inside the given ellipse, therefore each branch of hyperbola cuts the ellipse in two pints. Note that the distance of vertices of the hyperbola $(H)$ is $2m.$ The value $m$ increases when the vertices approach $w$ and $w^2$ becomes $m.$
Maximal value of $m$ is obtained when the hyperbola is degenerate (green line in the figure). In this case $2m=|w-w^2|.$ Only one point on the ellipse satisfies $$|z-w| - |z-w^2|=|w-w^2|=\sqrt3$$
The point on the ellipse is the green point in the figure.
Can you find its complex coordinate?
The points that maximize the difference of the distance will lie, by triangle inequality, on the line joining the two points i.e on $x =-\dfrac{1}{2}$. This line intersects the ellipse at $\left(-\dfrac{1}{2}, \dfrac{9\sqrt{11}}{10} \right)$ and at $\left(-\dfrac{1}{2}, -\dfrac{9\sqrt{11}}{10} \right)$
From this we readily obtain the difference of distances as $\sqrt 3$