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How to solve |ax-b|=0 when a,x,b are complex

  1. Nov 17, 2011 #1
    Hello,

    I have the following equation where a and b are complex constants, and x is a complex variable:

    [tex]\left\| a x - b\right\|^2=0[/tex]

    which can be rewritten as:

    [tex](ax-b)\overline{(ax-b)} = 0 [/tex]

    or alternatively:

    [tex]|a|^2 |x|^2 - 2\Re\{abx\} + |b|^2 = 0[/tex]

    How would you solve this equation for x?
    I set [itex]x=r e^{i\theta}[/itex], and tried to find values for r and θ that satisfy the equation, but it doesn't feel like a straightforward approach.

    Any hint?

    *** Note: *** the title of this thread contains a mistake and I cannot correct it now: I meant to write |ax-b|2
     
  2. jcsd
  3. Nov 17, 2011 #2

    Bacle2

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    Well, what is the only complex number who's modulus/length is identically zero?
     
  4. Nov 18, 2011 #3
    I suppose it was sufficient to impose ax-b=0, so that x=b/a ...
     
  5. Nov 19, 2011 #4

    Bacle2

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    Yes; that depends on the form of the solution that you are looking for; you could also use the fact that the real- and imaginary- parts should each equal zero. It depends on the format you want for the solution.
     
  6. Nov 19, 2011 #5

    Bacle2

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    I'm sorry, I did not read your question carefully-enough: I was suggesting to pre-multiply the
    initial expression az-b , separate into real- and imaginary- parts , find the modulus, and set each part equal to zero separately. This way you get a little more information on each of a,b,c. But I'm trying to think if there may be a more geometric way of finding a solution, i.e., if the points a,b, z satisfying your equation can be characterized as being part of some curve ( in the same way as , e.g., |z|=1 trivially describes a circle).
     
  7. Nov 19, 2011 #6
    Hi Bacle2,

    thanks for the explanations. Now it's pretty clear.

    I was thinking of possible "geometric interpretations" of the problem...I think that if you consider a simplified case like |z-b|=Δ, where Δ is a positive real number (we assumed a=1), then the solution of this equation are the points lying on a circle having center in b and radius Δ. Clearly, if we set Δ=0 the solution is z=b (the centre of the circle). Introducing the term a does not change the situation too much, in fact, the term az would then represent a rotation around the origin, plus a rescaling of the modulus of z.
     
  8. Nov 19, 2011 #7

    Bacle2

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    Good, nice job, mnb96 !.
     
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