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.
  • #1
mnb96
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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
 
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  • #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 !.
 

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

How do I solve |ax-b|=0 when a,x,b are complex?

To solve this equation, you first need to understand the concept of complex numbers. Complex numbers are numbers that consist of two parts - a real part and an imaginary part. In this equation, a, x, and b are all complex numbers. To solve for x, you will need to use the properties of complex numbers and the rules of algebra.

What is the process for solving |ax-b|=0?

The process for solving |ax-b|=0 involves isolating the variable x on one side of the equation. You can do this by using the properties of complex numbers and the rules of algebra, such as adding or subtracting the same number to both sides or multiplying or dividing both sides by the same number.

Can I use the quadratic formula to solve |ax-b|=0?

No, the quadratic formula is not applicable when dealing with complex numbers. Instead, you will need to use the properties of complex numbers and the rules of algebra to solve for x.

What are the solutions to |ax-b|=0?

The solutions to |ax-b|=0 will be complex numbers. There may be multiple solutions depending on the values of a and b. These solutions can be expressed in the form of a+bi, where a and b are real numbers and i is the imaginary unit.

Why is it important to solve |ax-b|=0 when working with complex numbers?

Solving |ax-b|=0 allows you to find the values of x that satisfy the equation and can help you solve other complex number equations. It also helps you understand the properties of complex numbers and how to manipulate them algebraically.

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