ellipse in complex numbers

Question

Argue geometrically that:

$$|z-4i| + |z+4i|=10$$ is an ellipse.

I've thought about it but have no good idea. I understand these are two distances, one to $z_o=4i$ and the conjugate, but what then? Would you give me a hand?

Answer 1

Hint:

Recall the definition of an ellipse (in terms of a locus of points). In defining an ellipse, we fix two points, called the foci, and then define the ellipse to be the set of all points such the sum of the distances from those foci is a given constant.

In effect, then, if we call the foci $F_1,F_2$ and arbitrary points $P$, and call our given constant $c$, then the ellipse is the set of points $P$ such that

$$|\overline{PF_1}| + |\overline{PF_2}| = c$$

holds true. Granted this construction is generally done in terms of $\Bbb R^2$ but it plays out largely similarly in $\Bbb C$ in this case.

Answer 2

Hint:

An ellipse is the locus of the points such that the sum of the distances to two foci is constant. This is the geometric interpretation.

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Analytically:

From $$\left|x+i\left(y+4\right)\right|+\left|x+i\left(y-4\right)\right|=10,$$ using the conjugate binomial, we have $$\left|x+i\left(y+4\right)\right|-\left|x+i\left(y-4\right)\right|=\frac{\left|x+i\left(y+4\right)\right|^2-\left|x+i\left(y-4\right)\right|^2}{10}=\frac{16y}{10}.$$

Then solving for the first term and squaring, $$x^2+\left(y+4\right)^2=\frac14\left(10+\frac{16y}{10}\right)^2$$

After simplification,

$$\frac{x^2}{9}+\frac{y^2}{25}=1.$$