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Loci of Complex Equation

  1. Jan 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the relation ## |\frac{z-i}{z*-i}| = \lambda ## where z = x + yi
    a) For ##\lambda = 1## show that the locus is a line in the complex plane and find its equation.
    b) What is the locus when ##\lambda = 0##?
    c) Show that for all other positive ##\lambda## the locus may be written as ##zz* + bz* + b*z + c = 0## where ##c## is a real number. Find conditions on complex #b# and real #c# for which the expression describes a
    i) Point
    ii) Circle
    iii) Line

    2. Relevant equations
    ## \frac{m}{n} = \frac{mn*}{|n|^2}##
    if ## m = k + ip ##, ##|m| = \sqrt{k^2+p^2}##

    3. The attempt at a solution
    I tried the following on the first part:
    ## 1 = |\frac{z-i}{z*-i}| = |\frac{x+i(y-1)}{x-i(y+1)}| = |\frac{(x+i(y-1))\cdot (x+i(y+1))}{x^2+(y+1)^2}| = |\frac{x^2-y^2-2xyi}{x^2+(y+1)^2}| ##
    but from here I am stuck, because this does not resemble the equation of a line at all.

    Thanks for all help and hints! :)
     
  2. jcsd
  3. Jan 15, 2017 #2

    Mark44

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    For the first part, with ##\lambda = 1##, you have ##|\frac{z-i}{\bar{z}-i}| = 1##, or equivalently, ##|z - i| = |\bar{z}-i|##. Simplify this equation by replacing z and ##\bar{z}##, and by taking magnitudes on each side.
     
  4. Jan 15, 2017 #3
    Right! That seems easier! So then we have:
    ## x^2 + (y-1)^2 = x^2 + (y+1)^2 ##
    ## y^2 -2y +1 = y^2 + 2y +1 ##
    ## y = 0##
    Thus the loci of all z must be the real axis?

    For part b) then: ##|z-i| = 0## requires z=i, thus the loci is a single point?

    in c) I have the following:
    ## \sqrt{(z-i)(z*+i)} = \lambda \sqrt{(z*-i)(z+i)} ## yielding:
    ##(\lambda^2 -1)z*z - (\lambda^2+1)i z + (\lambda^2+1)iz* + (\lambda^2-1) = 0 ##
    but I am not sure how I should derive the conditions for the different loci.
    Inserting ##z = x +yi## in ##z^*z + bz^* +b^*z + c = 0 ## and noting ##b = -b^*## and ##c=1## simply yields: ##x^2 + y^2 + 1 = 0## which is the equation of a parabola...
     
    Last edited: Jan 15, 2017
  5. Jan 15, 2017 #4

    Mark44

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    Yes on both a) and b).
    I haven't looked at c) yet, but will do so now.
     
  6. Jan 15, 2017 #5

    haruspex

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    Check that step
    Umm.. No it isn't.
     
  7. Jan 16, 2017 #6
    Well: ##z^*z + bz^* +b^*z + c = x^2+y^2+bx-by+b^*x+b^*y +c = 0 ##
    noting ##b = -b^*## and ##c=1## gives ##x^2+y^2+c=0##
    which is ##x^2+y^2+1=0##
    What am I doing wrong?
    Sorry, I meant hyperbola!
     
  8. Jan 16, 2017 #7

    SammyS

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    Why did you choose ##\ c=1\,?##
     
  9. Jan 16, 2017 #8

    haruspex

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    You are cancelling a zb with a z*b.
    Not that either.
     
  10. Jan 16, 2017 #9
    I didn't choose it, it followed from ##(\lambda^2 -1)z*z - (\lambda^2+1)i z + (\lambda^2+1)iz* + (\lambda^2-1) = 0 ## assuming ##\lambda \ne 1##
     
  11. Jan 16, 2017 #10
    Sorry, my bad!
    It should be: ##x^2 + y^2 - 2bi y + 1 = x^2 +y^2 + 2|b|y + 1 = 0## which can be written as ##x^2 + (y+|b|)^2 = |b|^2 - 1##. Which is the equation of a circle. This becomes a point if ##|b|^2 = 1##. But how can this be made a line?
     
  12. Jan 16, 2017 #11

    haruspex

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    To arrive at this equation you made an assumption about λ, remember?
     
  13. Jan 16, 2017 #12

    Ray Vickson

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    Use geometry! ##|z-i|## is the distance of the point ##(x,y)## from the point ##(0,1)## in the Cartesian plane. ##|z^*-i| = |(z^*-i)^*| = |z+i|## is the distance from ##(x,y)## to ##(0,-1)##.

    So, for (a), you want all the points ##(x,y)## that are the same distance to the two points ##(0,1)## and ##(0,-1)##.
    For (b) you want ##d[(x,y) \to (0,1)]= 0\, d[(x,y) \to (0,-1)] = 0##. For (c) you want ##d[(x,y) \to (0,1)] = \lambda d[(x,y) \to (0,-1)]##, or
    ##d[(x,y) \to (0,1)]^2 = \lambda^2 d[(x,y) \to (0,-1)]^2.## Expand this equation out in terms of powers of ##x## and ##y##, to see what you get.
     
  14. Jan 17, 2017 #13
    Yes, I assumed that ##\lambda >1##. This also means that I cannot insert ##lambda = 1## (which gave a line) into my ##z^*z + bz^* +b^*z + c = 0 ## equation, because in that case ##b## diverges. Therefore I am not really sure how to find the conditions of a line.
     
  15. Jan 17, 2017 #14

    haruspex

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    Sorry, I overlooked that it was asking for conditions on b and c, so is specifically in the context of λ≠1. As you found, it can never be a line, though it does tend to a line in the limit as b tends to infinity.
     
  16. Jan 18, 2017 #15
    Got it! Thank you very much!
     
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