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Plot $|z - i| + |z + i| = 16$ on the complex plane

Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.

I get to the point:

$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$

$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$

It seems like I'm and doing something the hard way.

  • 1
    The set of poi$n$ts with constant sum of distances from given two points is an [ellipse](http://en.wikipedia.org/wiki/Ellipse). A derivation of the equation of the ellipse, given the two points and the sum of distances, is shown e.g in [this question](http://math.stackexchange.com/questions/70732/why-do-definitions-of-distinct-conic-sections-produce-a-single-equation).2012-03-31

6 Answers 6

16

Maybe it's quicker way.

Equation $|z-i| + |z+i| = 16$ is equivalent to $ \sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2}. $ Squaring both sides you obtain $ x^2 + (y-1)^2 = 256 - 32\sqrt{x^2+(y+1)^2} + x^2 + (y+1)^2. $ Some terms cancel out hence you get $ 8\sqrt{x^2 + (y+1)^2} = 64 + y, $ and $ 64(x^2 + (y+1)^2) = 64^2 + 128y + y^2. $ Finally $ 64x^2 + 63y^2 = 64 \cdot 63. $ Now it is easy to write ellipse equation $ \frac{x^2}{63} + \frac{y^2}{64} = 1. $

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He asked for a plot:

enter image description here

from here, where I used $ \sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2} $ from the currently accepted answer.

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    ok, it is an ellipse and i revoke the -12012-04-05
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Just keep going: $2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$ $\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 127 - (x^2+y^2)$ $(x^2 + (y - 1)^2)(x^2 + (y + 1)^2) = (127 - (x^2+y^2))^2$ $x^4 + x^2[(y - 1)^2 + (y + 1)^2] + [(y+1)(y-1)]^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ (x^2+y^2)^2$ $x^4 + 2x^2y^2 +2x^2 + (y^2-1)^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$ $x^4 + 2x^2y^2 +2x^2 + y^4-2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$ $2x^2 -2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)$ $x^2(2+2\cdot127) +y^2(-2+2\cdot127) = 127^2 -1$ $256x^2 +252y^2 = 16128$ $\frac {x^2} {63} +\frac {y^2} {64} = 1$

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Don't we all love algebraic solutions...

The ellipse has an equation of the form

$(\frac {x} {a})^2 + (\frac {y} {b})^2 = 1$

The focals are aligned on the y axis, therefore

  • a is the side of a rectangle triangle, the other site being 1 and the hypothenuse 16/2:

$a = \sqrt{8^2-1}$

  • b is 16/2.

The equation of the ellipse is

$(\frac {x} {\sqrt{63}})^2 + (\frac {y} {8})^2 = 1$

Or, to write it as above

$\frac {x^2} {63} + \frac {y^2} {64} = 1$

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I think you are on the right track.

Wolframalpha also doesn't have a "simpler form" of the algebraic equation. (see http://www.wolframalpha.com/input/?i=%28%7Cx%2Biy-i%7C%2B%7Cx%2Biy%2Bi%7C%29%5E2%3D256)

Maybe this question is meant to be done on computer. Usually questions done by hand will ask you to "sketch" the graph.

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Sketching the graph gives a centrally symmetric ellipse with the imaginary axis as the major axis. The length of the half the major axis is clearly 8 (square 64), and pythagoras gives the square of half the minor axis as 63. This is reflected in the answer obtained by algebraic manipulation.

$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$

$\sqrt{(x^2 + y^2 + 1) - 2y}\sqrt{(x^2 + y^2 + 1) + 2y} = 127 - x^2 - y^2$

Square both sides:

$((x^2 + y^2 + 1) - 2y)((x^2 + y^2 + 1) + 2y) = (127 - x^2 - y^2)^2$

$(x^2+y^2+1)^2-4y^2 = 127^2+x^4+y^4-254x^2-254y^2+2x^2y^2$

Cancelling as we go and gathering terms in the obvious way:

$256x^2+252y^2=127^2-1=128 \times 126$

(difference of two squares)

The divisions now become easy and we reach the canonical form:

$\frac {x^2} {63} + \frac {y^2} {64} = 1$