sketch set satisfying $|z-2|+|z+2|\le5$ In the complex $z$ plane, $z = x+iy$, sketch the set satisfying the inequality: $|z-2|+|z+2|\le5$
I know from experience that this is an ellipse, but if I just wanted to find the $x$ and $y$ intercepts, is there a relatively quick way to do this, without bashing out a big mess of algebra?
Thanks everyone in advance!
 A: Set $y=0$ to get $|x-2|+|x+2|=5$ for the $x$=intercepts.
Corrected: The $y$-intercepts are a little messier, but as André pointed out in the comments, you can use the Pythagorean theorem to avoid most of the algebra. The $y$-intercepts at $\pm iy$ are equidistant from $2$ and $-2$, so for them we have $|iy-2|=|iy+2|=5/2$. Thus, $y^2+4=\frac{25}4$, $y^2=\frac94$, and $y=\pm\frac32$.
A: It depends on what you mean by a big mess of algebra. I understand that you want to find all points $z=x+iy$ with $x=0$ or $y=0$ respectively, which satisfy the inequality.
For $y=0$ this is rather simple, since you are just looking for the points on the real axis such that the distance to $2$ and $-2$ doesn't add up to more than $5$. This value is $4$ for all points lying between $2$ and $-2$, for values of $x>2$ the sum is $2(x-2)+4$ (and the case $x<2$ works similar).
For $x=0$ the distance to $2$ and $-2$ is equal, so you just need to compute $2|iy-2|\leq 5$. Expanding this gives you an easy condition on $y$.
A: If we take the two foci at $x=\pm 2$ and draw the ellipse such that the sum of the distances to the two foci is 5 (it is easy to find the cuts on the $Ox$ and $Oy$ axes), is the solution not the points $inside$ the ellipse then? Cheers! 
