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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 [itex]z_{1}[/itex], [itex]z_{2}[/itex], [itex]z_{3}[/itex] and [itex]z_{4}[/itex] the angles of [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}}[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}}[/itex] are the same, then the four numbers lie on a circle or on a straight line, and conversely.
[itex]x^{2} + y^{2} = r^{2}[/itex] and [itex]ax + b = c[/itex]?
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 = [itex]\rho(\cos(\phi) + i\sin(\phi))[/itex]. I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex].
Between two complex numbers, z and z', if the angle between the two are real, then:
[itex]\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'}}[/itex]
I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex] so that [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}= \frac{\rho_{123}}{\rho_{124}}[/itex].
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 [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}[/itex] to their [itex]a + bi[/itex] components to try to get them into the form for the equation of a circle? So that [itex]\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})}}[/itex]?
Does anyone have a hint for this problem?
Homework Statement
Prove that if for four complex numbers [itex]z_{1}[/itex], [itex]z_{2}[/itex], [itex]z_{3}[/itex] and [itex]z_{4}[/itex] the angles of [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}}[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}}[/itex] are the same, then the four numbers lie on a circle or on a straight line, and conversely.
Homework Equations
[itex]x^{2} + y^{2} = r^{2}[/itex] and [itex]ax + b = c[/itex]?
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 = [itex]\rho(\cos(\phi) + i\sin(\phi))[/itex]. I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex].
Between two complex numbers, z and z', if the angle between the two are real, then:
[itex]\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'}}[/itex]
I have [itex]\frac{z_{3} - z_{1}}{z_{3} - z_{2}} = \rho_{123}(\cos(\phi) + i \sin(\phi))[/itex] and [itex]\frac{z_{4} - z_{1}}{z_{4} - z_{2}} = \rho_{124}(\cos(\phi) + i \sin(\phi))[/itex] so that [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}= \frac{\rho_{123}}{\rho_{124}}[/itex].
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 [itex]\frac{\frac{z_{3} - z_{1}}{z_{3} - z_{2}}} {\frac{z_{4} - z_{1}}{z_{4} - z_{2}}}[/itex] to their [itex]a + bi[/itex] components to try to get them into the form for the equation of a circle? So that [itex]\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})}}[/itex]?
Does anyone have a hint for this problem?