Four complex numbers on a circle or on a straight line.

I'm reading R. Courant & H. Robbins' "What is Mathematics: An Elementary Approach to Ideas and Methods" for fun. I'm on page $$97$$ and of the second edition. There are $$8$$ exercises on complex numbers there, the seventh is as follows.

Exercise $$7$$: Prove that if for four complex numbers $$z_1, z_2, z_3, z_4$$ the angles of $$\alpha=\frac{z_3-z_1}{z_3-z_2}$$ and $$\beta=\frac{z_4-z_1}{z_4-z_2}$$ are the same, then the four numbers lie on a circle or on a straight line, and conversely.

My Attempt:

Lemma: The quotient of any two complex numbers with the same angle is real. This is Exercise $$6$$.

Proof: Let $$z=\rho_1(\cos(\phi)+i\sin(\phi))$$ and $$w=\rho_2(\cos(\phi)+i\sin(\phi))\in\mathbb{C}$$ have the same argument $$\phi$$, with $$\rho_1, \rho_2\in\mathbb{R}$$ and $$w\neq 0$$. Then $$\rho_2\neq 0$$ and \begin{align} \frac{z}{w}&=\frac{\rho_1(\cos(\phi)+i\sin(\phi))}{\rho_2(\cos(\phi)+i\sin(\phi))} \\ &=\frac{\rho_1}{\rho_2} \end{align} is real.$$\square$$

By the Lemma above, $$\frac{\alpha}{\beta}\in\mathbb{R}$$.

Where do I go from here?

Thoughts:

The Lemma seems much more applicable to Exercise 8, stated below for completeness.

Exercise 8: Prove that four points lie on a circle or on a straight line if and only if $$\frac{\alpha}{\beta}\in\mathbb{R}$$.

• Anything omitted? the angles of $\alpha=\frac{z_3-z_1}{z_3-z_2}$ and $\beta=\frac{z_4-z_1}{z_4-z_2}$, Commented Feb 7, 2017 at 11:41
• @MyGlasses Yes, "are the same"; thank you :) Commented Feb 7, 2017 at 11:54

First, let's get a geometric interpretation of the statement "the angles of $\alpha$ and $\beta$ are the same".

That's a picture of $z_1$, $z_2$, and $z_3$. Note that $\theta_1$ and $\theta_2$ represent the angles of $z_3 - z_1$ and $z_3 - z_2$. Therefore, the angle of $\alpha$ (which is $\theta_1 - \theta_2$) is the measure of the geometric angle $z_1z_3z_2$.

So rephrasing the problem in geometric terms, we have:

Suppose four points $A$, $B$, $C$, and $D$ are in the plane such that the angles $BCA$ and $ADB$ are equal. Prove that all four points are in a straight line or lie on a common circle. Conversely, given four points $A$, $B$, $C$, and $D$ either on a line or all on a circle, prove that the angles $ACB$ and $ADB$ are equal.

Where we add the restriction that's unusual for geometry - but is inherited from the complex plane - that an angle $XYZ$ is measured counter-clockwise from $YZ$ to $YX$. (so that the measure of the angle $XYZ$ is $2\pi$ radians minus the measure of the angle $ZYX$) We also allow two angles to be considered equal if their measures differ by exactly $\pi$ radians, to reflect the idea that $z$ and $-z$ have the same angle.

First let's tackle the direction "on a line or a circle implies equal angles". Now, if the four points lie on a line then obviously the angles $BCA=ACB$ and $ADB$ are equal (since they're both $0$ radians or $\pi$ radians).

If the four points lie on a circle, then:

This shows $A$, $B$, and $C$ all on a circle, and two different possibilities for point $D$: either it lies on the same side of the line $\overline{AB}$ as point $C$ (case $D_2$) or it lies on the opposite side. (case $D_1$)

By the central angle theorem, $\theta_1 = {1 \over 2} \phi = \theta_2$, so if $D$ is in case $D_2$ we have the angles equal. If $D$ is in case $D_2$, we have that \begin{align} 2\pi - \theta_3 &= {1 \over 2} (2\pi - \phi) \\ &= \pi - {1 \over 2} \phi \\ &= \pi - \theta_1 \end{align}

And therefore $\theta_3 = \theta_1 + \pi$. (which we were going to consider equal)

For the other direction, first take the case that both angles are $0$ or $\pi$. If that's the case, then $C$ and $D$ must both line on line $\overline{AB}$ and we're done. (this corresponds to $\alpha$ and $\beta$ both being real numbers)

Now, if we have some other angle value, then $A$, $B$, and $C$ do not all lie on a line, and therefore there exists a unique circle $O_1$ containing $A$, $B$, and $C$. Likewise, there is a unique circle containing points $A$, $B$, and $D$. Since the central angle $AO_1B$ is equal to the central angle $AO_2B$ (both are twice the angle $ACB$), the radii of the two circles are identical, and from there it's just a bit of case analysis (taking the case where $C$ and $D$ are on the same side of $\overline{AB}$ or on opposite sides, and whether the measures of $ACB$ and $ADB$ are really equal or are off by $\pi$ radians) to show that both circles are identical.

I don't know if you would count this as proof, but here's the idea:

Angles subtended on a circle by the same arc on the same side are equal. Since the converse is also true, if $$z_3$$ and $$z_4$$ do not lie anywhere on the line joining $$z_1$$ and $$z_2$$, i.e. $$\alpha$$ and $$\beta$$ are not purely real, then $$z_1$$, $$z_2$$, $$z_3$$ and $$z_4$$ lie on a circle.

If $$\alpha =\frac{z_3-z_1}{z_3-z_2}$$ and $$\beta= \frac{z_4-z_1}{z_4-z_2}$$ are both purely real, then $$z_3$$ and $$z_4$$ lie on the line joining $$z_1$$ and $$z_2$$.

Combining this with the fact that $$\frac{\alpha}{\beta} \in \Bbb{R}$$ also solves the problem posed in Exercise 8, I guess.