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Give a geometric description of the following set: $$\{z:|z-2|+|z+2|=5\}$$

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What did you try ? – Belgi Jan 8 '13 at 21:48
What is the set of those points, for which sum of distances from two fixed points is constant $(=5)$? – M. Strochyk Jan 8 '13 at 22:00
up vote 3 down vote accepted

You just have to read what you wrote, if we have a complex number $w$ then we can give a geometric meaning for $|w-5|$ as the (euclidean) distance from the complex number $w$ to the complex number $5$.

With that in mind we can $\textit{read}$ your set as:

those complex numbers $z$ such the distance from $z$ to $2$ added to the distance from $z$ to $-2$ is $5$.

That is, a set of points such the sum of the distances from each of those points to two fixed points is constant. And that is the definition of an ellipse whose focuses (in this case) are $2$ and $-2$.

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But the sum of the distances is not constant, it's 5. – i_a_n Jan 8 '13 at 22:19
@i_a_n: 5 is a constant! Also, Zango,since this is HW, you shouldn't give a complete solution. – Grumpy Parsnip Jan 8 '13 at 22:19
Sorry! I will try to be more "hinty" next time and not give the complete solution! – Z. L. Jan 8 '13 at 22:24

You can think of complex numbers as vectors in a two-dimensional Cartesian plane, so that you have the following:

$\{(x,y) : |(x,y) - (2,0)| + |(x,y) + (2,0)| = (5,0) \}$

This is basically saying that the sum of differences between any valid point $(x,y)$ from $(2,0)$ and $(-2, 0)$ must be equal to 5. Thus, the sum of the two line segments from $(2,0)$ to $(x,y)$ to $(-2,0)$ must be of length 5.

If you want to visualize this, imagine you have a belt of length 5 attached to those two points. Then, using a pencil (or other stick), stretch the belt to its maximum length and move the pencil around in the plane. Which points does it pass through?

Then, describe this set numerically. First consider where it intersects the $x$ and $y$ axes and then the points in between.


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