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

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.

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The set of points with constant sum of distances from given two points is an 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. –  Martin Sleziak Mar 31 '12 at 9:51

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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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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177 should be 127 –  Mark Bennet Mar 31 '12 at 9:52
@MarkBennet, indeed. Thanks. –  davin Mar 31 '12 at 9:59

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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-1 because this is not the plot of the complex equation of the question –  miracle173 Mar 31 '12 at 11:48
@miracle173, why? –  draks ... Apr 5 '12 at 7:34
This is $|z|=8$ –  Emre Apr 5 '12 at 7:51
@Emre it's not, it's an ellipse (with a flattening factor $g=1-\frac {b}{a}=\frac{1}{64}$)! See the link(tinyurl.com/clunqaw)! –  draks ... Apr 5 '12 at 10:13
ok, it is an ellipse and i revoke the -1 –  miracle173 Apr 5 '12 at 21:02

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:

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

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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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