QuantumP7
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This isn't really homework. I'm studying What Is Mathematics by myself. But I'm very stuck on one of its exercises.
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.
x^{2} + y^{2} = r^{2} and ax + b = c?
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?
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?