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