Four complex points lie on a circle proof

[QuantumP7]

This isn't really homework. I'm studying What Is Mathematics by myself. But I'm very stuck on one of its exercises.

Homework Statement

Prove that if for four complex numbers z_{1}, z_{2}, z_{3} and z_{4} the angles of \frac{z_{3} - z_{1}}{z_{3} - z_{2}} and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} are the same, then the four numbers lie on a circle or on a straight line, and conversely.

Homework Equations

x^{2} + y^{2} = r^{2} and ax + b = c?

The Attempt at a Solution

I can prove that the quotient between two complex numbers with the same angle is real, using the trigonometric form of the complex numbers (z = \rho(\cos(\phi) + i\sin(\phi)). I have \frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi)) and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi)).

Between two complex numbers, z and z', if the angle between the two are real, then:

\frac{z}{z'} = \frac{\rho_{z}(cos(\phi) + i \sin(\phi))}{\rho_{z'}(\cos(\phi) + i \sin(\phi))} = \frac{\rho_{z}}{\rho_{z'}}(\cos(\phi - \phi) + i \sin(\phi - \phi)) = \frac{\rho_{z}}{\rho_{z'}}(\cos{0} + i \sin {0}) = \frac{\rho_{z}}{\rho_{z'}}

I have \frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi)) and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi)) so that \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}= \frac{\rho_{123}}{\rho_{124}}.

What I am not understanding is how this means that all four complex points lie on a circle or a straight line? Should I reduce the \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}} to their a + bi components to try to get them into the form for the equation of a circle? So that \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}} = \frac{\frac{(a_{3} + ib_{3}) - (a_{1} + ib_{1})}{(a_{3} + ib{3}) - (a_{2} + ib{2})}}{\frac{(a_{4} + ib_{4}) - (a_{1} + ib_{1})}{(a_{4} + ib_{4}) - (a_{2} + ib_{2})}}?

Does anyone have a hint for this problem?

 

[Dick]

This isn't really homework. I'm studying What Is Mathematics by myself. But I'm very stuck on one of its exercises.

Homework Statement

Prove that if for four complex numbers z_{1}, z_{2}, z_{3} and z_{4} the angles of \frac{z_{3} - z_{1}}{z_{3} - z_{2}} and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} are the same, then the four numbers lie on a circle or on a straight line, and conversely.

Homework Equations

x^{2} + y^{2} = r^{2} and ax + b = c?

The Attempt at a Solution

I can prove that the quotient between two complex numbers with the same angle is real, using the trigonometric form of the complex numbers (z = \rho(\cos(\phi) + i\sin(\phi)). I have \frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi)) and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi)).

Between two complex numbers, z and z', if the angle between the two are real, then:

\frac{z}{z'} = \frac{\rho_{z}(cos(\phi) + i \sin(\phi))}{\rho_{z'}(\cos(\phi) + i \sin(\phi))} = \frac{\rho_{z}}{\rho_{z'}}(\cos(\phi - \phi) + i \sin(\phi - \phi)) = \frac{\rho_{z}}{\rho_{z'}}(\cos{0} + i \sin {0}) = \frac{\rho_{z}}{\rho_{z'}}

I have \frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi)) and \frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi)) so that \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}= \frac{\rho_{123}}{\rho_{124}}.

What I am not understanding is how this means that all four complex points lie on a circle or a straight line? Should I reduce the \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}} to their a + bi components to try to get them into the form for the equation of a circle? So that \frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}} = \frac{\frac{(a_{3} + ib_{3}) - (a_{1} + ib_{1})}{(a_{3} + ib{3}) - (a_{2} + ib{2})}}{\frac{(a_{4} + ib_{4}) - (a_{1} + ib_{1})}{(a_{4} + ib_{4}) - (a_{2} + ib_{2})}}?

Does anyone have a hint for this problem?

Use the angle being the same to show the cross ratio is real. The cross ratio is invariant under Mobius transformations. Use ideas like that rather than trying to mess around with components.

 

[pasmith]

I think this is a question about Mobius maps. I say this because
\phi(z) = \frac{z - z_1}{z - z_2}
is a Mobius map, and one of the properties of Mobius maps is that they map straight lines and circles to straight lines and circles. Now
\phi^{-1}(z) = \frac{zz_2 - z_1}{z - 1}
maps the points 0, \infty, (z_3 - z_1)/(z_3 - z_2) and (z_4 - z_1)/(z_4 - z_2) to the points z_1, z_2, z_3, z_4 respectively. Now if
<br /> \arg \left( \frac{z_3 - z_1}{z_3 - z_2} \right) = \arg \left( \frac{z_4 - z_1}{z_4 - z_2} \right)<br />
then the points 0, \infty, (z_3 - z_1)/(z_3 - z_2) and (z_4 - z_1)/(z_4 - z_2) lie on a straight line, so their images lie on a straight line or a circle.

Conversely, if z_1, z_2, z_3 and z_4 lie on a straight line or a circle, then so do 0, \infty, (z_3 - z_1)/(z_3 - z_2) and (z_4 - z_1)/(z_4 - z_2). Since \infty is in that set, it must be a straight line (so that the arguments of (z_3 - z_1)/(z_3 - z_2) and (z_4 - z_1)/(z_4 - z_2) are equal or differ by \pi).

 

[Dick]

There's an even simpler argument if you've already proven that given three distinct points there is some mobius transformation such that f(z1)=1, f(z2)=0 and f(z3)=∞. Then you don't really have to worry about the details of what the formula for f even is. Just use that the cross ratio is real.

 

[QuantumP7]

Thank you guys soooooooooo much!