TITLE: If $0$, $z_1$, $z_2$ and $z_3$ are concyclic, then $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear
QUESTION [8 upvotes]: If the complex numbers $0$, $z_1$, $z_2$ and $z_3$ are concyclic, prove that $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear.

I really can't seem to get anywhere on this problem, but all I've deduced is that there might be some relationship between circle geometry properties and the arguments of the complex numbers.

REPLY [2 votes]: The "mature" way to think of this is as a function on the Riemann sphere - this the standard complex plane with a point $\infty$ added, also called the "extended complex plane". The function $1/z$ is a special case of what is called a linear fractional transformation - these are all functions of the form $\frac{az+b}{cz+d}$, where the coefficients are in $\mathbb{C}$. One property of linear fractional transformations is that they send "generalized circles" to "generalized circles", where a generalized circle is a either a normal circle or a line (a line is a called a generalized circle because you can view it as a circle through the point at infinity).
In this case, we see our circle through 0 will be sent through a "circle through infinity" because we define $1/0$ to be $\infty$ here. So your four points will be sent to a line, and in particular, $1/z_1$, $1/z_2$, and $1/z_3$ will be collinear. 
Linear fractional transformations and their properties are a standard topic in complex analysis, and you don't even need to know any analysis to prove the property I just mentioned. Any standard text on complex analysis should have the proof. I suggest you find a copy of Gamelin's book, if you are interested.