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

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In summary, the equation has a solution if and only if b is the only complex number with a modulus of zero.f
  • #1
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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
Well, what is the only complex number who's modulus/length is identically zero?
 
  • #3
I suppose it was sufficient to impose ax-b=0, so that x=b/a ...
 
  • #4
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.
 
  • #5
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).
 
  • #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.
 
  • #7
Good, nice job, mnb96 !.
 

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