Complex analysis graphing an ellipse. Give a geometric argument that $|z-4i|+|z+4i|=10$.
My attempt: $|z-4i|$ represents the distance from an arbitrary point to the coordinate $(0,4i)$ while $|z+4i|$ is the distance from an arbitrary point to the coordinate $(0,-4i)$.
Then stuck; I don't know how this relates to an ellipse...Thanks for the help.
 A: An ellipse is a set of points such that the sum of the distances from each point in the set to a pair of fixed points is a constant.
Here is a derivation of the equation of an ellipse from this definition:
Imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$.  We draw an ellipse by holding a pen taut against the string.  The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable.  First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic.  Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$.  We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
