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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!

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3 Answers 3

up vote 2 down vote accepted

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 $y=\pm\frac32$.

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Hi Dr. Scott, I got the x = 5/2 (plus and minus) and y = 3/2 (plus and minus). Is this right? Solving for x when y = 0 was easier for me, did you mean the 1st one was trivial? –  JackReacher May 31 '12 at 11:47
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For the $y$-intercept, by symmetry want distance from $(0,y)$ to $(2,0)$ to be $5/2$. By Pythagorean Theorem, get $y^2+4=25/4$. –  André Nicolas May 31 '12 at 12:19
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@mathstudent: I should have gone to bed an hour ago. You’re quite right, as André points out in his comment, and I’ve corrected the answer accordingly. –  Brian M. Scott May 31 '12 at 12:27

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 $y$.

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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!

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