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

[mnb96]

Hello,

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

\left\| a x - b\right\|^2=0

which can be rewritten as:

(ax-b)\overline{(ax-b)} = 0

or alternatively:

|a|^2 |x|^2 - 2\Re\{abx\} + |b|^2 = 0

How would you solve this equation for x?
I set x=r e^{i\theta}, 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

 

[Bacle2]

Well, what is the only complex number who's modulus/length is identically zero?

 

[mnb96]

I suppose it was sufficient to impose ax-b=0, so that x=b/a ...

 

[Bacle2]

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.

 

[Bacle2]

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).

 

[mnb96]

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.

 

[Bacle2]

Good, nice job, mnb96 !.