# Complex line

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• Oct 10th 2012, 02:25 PM
eulcer
Complex line
Quick question, why does the equation z=(a+b)-a*b*conjugate(z) determine the line that passes through points a and b on the unit circle in the complex plane?
Thanks,
• Oct 10th 2012, 03:28 PM
MaxJasper
Re: Complex line
$\displaystyle y = \frac{i (a+b-x-a b x)}{-1+a b}$
• Oct 10th 2012, 03:51 PM
johnsomeone
Re: Complex line
I think you want it to be z = (a+b) - abz* (where w* means w-conjugate)

Assume z satisfies that equation. Write z = x + iy, where x and y are real. When you multiply everything out, you'll get 2 equations (one for the real part, one for the imaginary part) in two unknowns. Usually, there's a unique solution for x and y (hence for z), but if the numbers are just right, there can be no solution, or an infinite number of solutions. The later must be the case here, if, as the problem claims, that equation determines a line. What's happened is that the 2 equations you get are in fact the same equation, except for some multiplier. When that happens, you have one equation for x and y, and so will have a line of z's satifying the original equation.

With that in mind, it's possible to get to the answer without going through the work of setting z = x+iy and multiplyinbg everything out, and solving the syetem of equations given by the real and imaginary parts.

The usual solution to that equation is a single value for z, but sometimes it has no solution, and sometimes it has a line as a solution. Now, *if* you can show two distinct points satisfy the original equation, then it must be the line option. So if you can find two points satisfying the equation, that suffices to prove that it's a line between those two points.

Actually, must also exclude the uber-degenerate case that the solution set is the entire plane - so find any one point that doesn't satisfy the equation. Check that now:
If a+b not 0, then z = 0 is not a solution, proving the solution set isn't the whole plane.
If a + b = 0, then it becomes z = a^2 z*.
Check z = a+1: LHS = a + 1. RHS = a^2 (a*+1) = a(aa*) + a^2 = a+ a^2. Thus LHS = RHS only if a in {- 1, 1}.
So a in {- 1, 1} and a+b=0 remains as the only way it could be degenerate to the whole plane. But in that case it becomes z = a^2z* = z*, and z = i is not a solution to that.
Thus for no a and b on the unit circle does that equation degenerate to be the whole plane.

Now:
Plug in z = a. Get LHS = z = a. RHS = (a+b)-abz* = a+(b-ab(a)*) = a+b(1-(aa*)) = a + b(1-1) = a + b(0) = a = LHS. Since aa* = 1.
Thus z = a is a solution to the equation.

Plug in z = b. Get LHS = z = b. RHS = (a+b)-abz* = b+(a-ab(b)*) = b+a(1-(bb*)) = b + a(1-1) = b + a(0) = b = LHS. Since bb* = 1.
Thus z = b is a solution to the equation.

Thus the equation z = (a+b) - abz* either has no solution, or is a point, or a line, or is the whole plane.
Have shown if a and b are on the unit circle, then z=a and z = b are solutions to that equation.
Have also shown that the solution set isn't the whole plane.
Therefore, if a and b are distinct, then the solution set must be a line through a and b.
If a = b, then the solution set is either {a}, or it's a line through a.
• Oct 10th 2012, 03:54 PM
johnsomeone
Re: Complex line
If by that you intend that z = x+iy, I think you're overlooking that we're to assume that a and b are complex, and so that y value might not be real.
• Oct 10th 2012, 04:38 PM
eulcer
Re: Complex line
Quote:

Originally Posted by johnsomeone
I think you want it to be z = (a+b) - abz* (where w* means w-conjugate)

Assume z satisfies that equation. Write z = x + iy, where x and y are real. When you multiply everything out, you'll get 2 equations (one for the real part, one for the imaginary part) in two unknowns. Usually, there's a unique solution for x and y (hence for z), but if the numbers are just right, there can be no solution, or an infinite number of solutions. The later must be the case here, if, as the problem claims, that equation determines a line. What's happened is that the 2 equations you get are in fact the same equation, except for some multiplier. When that happens, you have one equation for x and y, and so will have a line of z's satifying the original equation.

With that in mind, it's possible to get to the answer without going through the work of setting z = x+iy and multiplyinbg everything out, and solving the syetem of equations given by the real and imaginary parts.

The usual solution to that equation is a single value for z, but sometimes it has no solution, and sometimes it has a line as a solution. Now, *if* you can show two distinct points satisfy the original equation, then it must be the line option. So if you can find two points satisfying the equation, that suffices to prove that it's a line between those two points.

Actually, must also exclude the uber-degenerate case that the solution set is the entire plane - so find any one point that doesn't satisfy the equation. Check that now:
If a+b not 0, then z = 0 is not a solution, proving the solution set isn't the whole plane.
If a + b = 0, then it becomes z = a^2 z*.
Check z = a+1: LHS = a + 1. RHS = a^2 (a*+1) = a(aa*) + a^2 = a+ a^2. Thus LHS = RHS only if a in {- 1, 1}.
So a in {- 1, 1} and a+b=0 remains as the only way it could be degenerate to the whole plane. But in that case it becomes z = a^2z* = z*, and z = i is not a solution to that.
Thus for no a and b on the unit circle does that equation degenerate to be the whole plane.

Now:
Plug in z = a. Get LHS = z = a. RHS = (a+b)-abz* = a+(b-ab(a)*) = a+b(1-(aa*)) = a + b(1-1) = a + b(0) = a = LHS. Since aa* = 1.
Thus z = a is a solution to the equation.

Plug in z = b. Get LHS = z = b. RHS = (a+b)-abz* = b+(a-ab(b)*) = b+a(1-(bb*)) = b + a(1-1) = b + a(0) = b = LHS. Since bb* = 1.
Thus z = b is a solution to the equation.

Thus the equation z = (a+b) - abz* either has no solution, or is a point, or a line, or is the whole plane.
Have shown if a and b are on the unit circle, then z=a and z = b are solutions to that equation.
Have also shown that the solution set isn't the whole plane.
Therefore, if a and b are distinct, then the solution set must be a line through a and b.
If a = b, then the solution set is either {a}, or it's a line through a.

I didn't follow that completely.
"The usual solution to that equation is a single value for z, but sometimes it has no solution, and sometimes it has a line as a solution."
Why is this the case?
• Oct 11th 2012, 08:05 AM
johnsomeone
Re: Complex line
2 LINEAR equations in 2 unknowns. Here are the only things that can happen:
-----------------------------
x + 2y = 3
2x +4y = 8
Degenerate. No solutions.
-----------------------------
x + y = 2x +4
x - y = -4
One solution: x = 1, y = 5.
That's the typical case.
-----------------------------
x + y = 4
-6x -3y = -3x - 12
Degenerate - one equation is a multiple of the other.
Solution is the line y = -x +4
-----------------------------
2(x-y)+3 = 4(x+y+1)-2x-6y-1
2x-y+1 = 2(x+y+2) - 3(y+2) +3
Uber-degenerate - since both equations amount to 0 = 0.
Solution is the entire plane, since every value of x and y satisfy both those equations.
-----------------------------