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Showing that the solutions form an ellipse

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Let |z+1| + |z-1| = 7 where z are complex numbers. Show that the solutions to this equation form an ellipse with foci at (+/-)1

    2. Relevant equations
    (x^2 / a) + (y^2 / b) = 1 equation for an ellipse

    3. The attempt at a solution
    I set z = a + bi and so |z-1| = ((a-1)^2 + b^2)^1/2 and analogously for |z+1|. I'm a bit confused now, if I square both sides I wind up with a big mess, how to I beat this into a form that's recognizable as an ellipse?
     
  2. jcsd
  3. Jan 29, 2017 #2

    Ray Vickson

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    Keep expanding: if ##d_1=|z-i|## and ##d_2 = |z+i|##, you have ##d_1+d_2 = 7##, so ##2 d_1 d_2 = 49 - d_1^2 - d_2^2##. Square both sides again and see what you get.
     
  4. Jan 30, 2017 #3
    but d_1 and d_2 are |z (+/-) 1| not |z(+/-)i|
     
  5. Jan 30, 2017 #4

    Ray Vickson

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    So, make the necessary changes. It really does not matter.
     
  6. Jan 30, 2017 #5
    I'm not sure if this is as neat as you make it look, I mean squaring the d_1^2 and d_2^2 is a messy ordeal in itself. I need to extract the real and complex parts from the d_1 and d_2 before i beat it into elliptical form, right?
     
  7. Jan 30, 2017 #6
    Squared it again and got 2401 - 98(d_1)^2 - 98(d_2)^2 - 2(d_1*d_2)^2 + d_1^4 + d_2^4 = 0

    These d^4 are scaring me, this looks like it's just getting more messy, am I doing this right?
     
  8. Jan 30, 2017 #7

    Ray Vickson

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    Expand it out in detail. You cannot predict what will happen if you do not try it. In other words, ##d_1^2 = (x-1)^2+y^2 = x^2 - 2x + 1 + y^2##, and ##d_1^4 = (x^2+y^2-2x+1)^2 = \cdots ##, etc.
     
  9. Jan 30, 2017 #8

    haruspex

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    Not sure that it is any easier, but you can limit the mess by working as far as possible with ##z## and ##\overline z##. First, derive an expansion for |z-1|2.
     
  10. Jan 30, 2017 #9

    Ray Vickson

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    Actually, instead of what I suggested in #7, write ##2 d_1 d_2 = 7^2 - d_1^2 - d_2^2##, and write out the right-hand side here as ##A x^2 + B y^2 + Cx+D## before squaring it. That makes things a lot easier----still messy, but easier.
     
  11. Jan 31, 2017 #10

    Ray Vickson

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    You are on the wrong track, We have ##|z - (a + bi)| = \sqrt{(x-a)^2+(y-b)^2}##, so it is not complicated at all.
     
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