Complex Ellipse to Cartesian Form

Question

I have a problem, where the equation of my ellipse is:
$$1=\frac{x^2}{75}+\frac{y^2}{56.25}$$
and I have been able to translate the equation to a complex form of
$$|z+2.5\sqrt{3}|+|z-2.5\sqrt{3}|=10\sqrt{3}.$$
Now the problem is being able to solve and show the method of substituting and rearranging from the complex form to the cartesian form. I know the $z$ has to be substituted with $x+yi$ but the rest is unknown. Please explain?

Answer 1

Given the ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

there are several ways to express it in the complex plane. Some are shown in the other answers. Here are two direct expressions for $z(\theta),~\theta\in[0,2\pi]$.

$$ z=a\cos\theta+ib\sin\theta $$ $$ r=\frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}} $$ $$ z=re^{i\theta} $$

Answer 2

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be an equation of the ellipse.

We need $$|z-c|+|z+c|=2a$$

Thus, $$c^2=a^2-b^2$$ or $$c^2=18.75,$$ which gives $c=2.5\sqrt3$, $a=5\sqrt3$ and since in the Gauss plain $z=x+yi$ it's just the point $(x,y)$,

we get the answer: $$|z-2.5\sqrt3|+|z+2.5\sqrt3|=10\sqrt3.$$ Done!

P.S. $|z-2.5\sqrt3|$ it's the distance between $M(x,y)$ and $F_1(2.5\sqrt3,0)$;

$|z+2.5\sqrt3|$ it's the distance between $M(x,y)$ and $F_2(-2.5\sqrt3,0)$.

By the ellipse definition $$MF_1+MF_2=constant,$$ where $F_1$ and $F_2$ are different points and our $constant>0.$

In our case the $constant=2a=10\sqrt3.$

Also, we have $$|z-2.5\sqrt3|=|x-2.5\sqrt{3}+yi|=\sqrt{(x-2.5\sqrt3)^2+y^2}=MF_1,$$ where $M(x,y)$ and $F_1(2.5\sqrt3,0)$.