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If we substitute $z=x+iy$ to $\lvert z-2\rvert + \lvert z+2\rvert=5$ and solve for $iy$ we will get

$$ iy=-\frac{1}{2}(2x-5) $$ $$ iy=-\frac{1}{2}(2x+5) $$

Then we can draw like in the $\mathbb{R}^2$ plane. This way we get two parallel lines.

Do you think this approach is correct for the complex plane? Just want to make sure.

  • 1
    You seem to have ignored the modulus signs... what you get by substituting $z=x+iy$ is $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$. – 2017-01-09
  • 0
    `solve for iy` The next two equations are wrong and don't follow from the given relation. – 2017-01-09

4 Answers 4

0

the equation is given by $$\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$$ square it!

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    Actually I did, but got a massive square root "in the middle". – 2017-01-09
  • 0
    no, you must square two times – 2017-01-09
  • 0
    Right, but then we get $2$ times two square roots "in the middle" on LHS. – 2017-01-09
  • 0
    Unenlightening answer, as the comments show. – 2017-01-10
6

This is an ellipse with foci at $2$ and $-2$.

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    So it is centered at the origin and symmetrical with the axes. Put $x=0$ and $y=0$ in the eq. in*@kccu*'s comment to find the semi-axes. – 2017-01-09
2

$1)$ If you want go through a geometric approach: by definition of ellipse one can conclude that $z$ is on an ellipse with foci $-2$ and $2$ and bigger axis equal to $5$.

$2)$ If you want go through an analytic approach you can write $z=x+iy$ and then

$$\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$$

$$\sqrt{(x-2)^2+y^2}=5-\sqrt{(x+2)^2+y^2}$$

Square both sides:

$$(x-2)^2+y^2=25-10\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2$$

$$10\sqrt{(x+2)^2+y^2}=25+8x \rightarrow 100[(x+2)^2+y^2]=(25+8x)^2$$

Can you finish?

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    How do we draw the ellipse exactly form a bunch of square roots? We can square LHS and RHS, but we'll still get one square root. How do we deal with it for precise drawing? – 2017-01-09
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    take a look again. Is that clear? You should be able to finish. – 2017-01-09
1

You are looking for the set of points, for which the sum of distances from points $p_0=2$ and $p_1=-2$ is equal to $5$.

Sounds like an ellipse.

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    But if -2 and 2 are two foci of the ellipse then it doesn't looks like the distance from -2 to 2 which equals 5. – 2017-01-09
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    @sequence - foci don't have to belong to an ellipse. I've changed my answer a little to be clearer. – 2017-01-09