Given four complex numbers $z_1, z_2, z_3$ and $z_4,$ show that they lie on a circle.

Question

Given four complex numbers $z_1, z_2, z_3$, and $z_4,$ show that they lie on a circle if $$\arg\left(\frac{z_4-z_1}{z_4-z_2}\right)=\arg\left(\frac{z_3-z_1}{z_3-z_2}\right).$$ How can I interpret this equality? And how can I show this statement? I need some help. Thanks.

Answer 1

Think of $z_1, z_2, z_3, z_4$ as four points on the plane with $z_1$ and $z_2$ the base points of two angles with summits $z_3$, $z_4$. Then the condition states that the angles subtended by $z_4$ and $z_3$ from $z_1$ and $z_2$ are equal. This is the condition in elementary geometry for four points to be co-cyclic. Since the angle subtended by a point on the circle is half the angle subtended by the centre.

Answer 2

First of all,

$$\arg\left(\frac{z_4-z_1}{z_4-z_2}\right)$$

represents the angle $\angle Z_2 Z_4 Z_1$. Similarly

$$\arg\left(\frac{z_3-z_1}{z_3-z_2}\right)$$

represents the angle $\angle Z_2 Z_3 Z_1$. This is because to when you divide two complex numbers, you subtract the angles and we don't care about the magnitudes. Euler's form for complex numbers makes it a bit clearer:

$$M_1 e^ {\theta_1} \div M_2 e^ {\theta_2} = \frac{M_1}{M_2}e^{\theta_1 - \theta_2}$$

Combine that with:

http://en.wikipedia.org/wiki/Inscribed_angle_theorem#Theorem

The $A$ and $B$ in the wikipedia graphic are your $z_1$ and $z_2$.

Answer 3

For an algebraic proof you can do it this way. Suppose that $${\rm arg}\Bigl(\frac{z-z_1}{z-z_2}\Bigr)=\alpha\ .\tag{$*$}$$ Let the modulus of the bracketed term be $r$; then we have $$\frac{z-z_1}{z-z_2}=re^{i\alpha}\ .$$ Multiplying out and taking conjugates, $$z-z_1=(z-z_2)re^{i\alpha}\quad\hbox{and}\quad \overline z-\overline{z_1}=(\overline z-\overline{z_2})re^{-i\alpha}\ .$$ Eliminating $r$ from these equations gives $$(z-z_1)(\overline z-\overline{z_2})e^{-i\alpha} =(\overline z-\overline{z_1})(z-z_2)e^{i\alpha}\ ;$$ collecting similar terms, $$z\overline z(e^{-i\alpha}-e^{i\alpha}) -z(\overline{z_2}e^{-i\alpha}-\overline{z_1}e^{i\alpha}) -\overline z(z_1e^{-i\alpha}-z_2e^{i\alpha}) =z_1\overline{z_2}e^{-i\alpha}-\overline{z_1}z_2e^{i\alpha}.$$ Now if $\alpha=n\pi$ then $(*)$ is the equation of a line; from now on assume $\alpha\ne n\pi$. Then $e^{-i\alpha}-e^{i\alpha}$ is not zero. If we write $$c=\frac{z_2e^{i\alpha}-z_1e^{-i\alpha}}{e^{i\alpha}-e^{-i\alpha}} \quad\hbox{and}\quad R=\left|\frac{z_1-z_2}{e^{i\alpha}-e^{-i\alpha}}\right|\ ,$$ then $R$ is a positive real number (assuming $z_1\ne z_2$), and after a bit of a struggle we can show that the previous equation becomes $$z\overline z-z\overline c-\overline zc+c\overline c=R\ ,$$ in other words $$|z-c|=\sqrt R\ .$$ This is the equation of the circle with centre $c$, passing through $z_1$ and $z_2$; since your $z_3$ and $z_4$ both satisfy $(*)$, they both lie on this circle.