# 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

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.

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There is a very close relationship between circle geometry and complex numbers. Inversion of complex numbers geometrically (almost) corresponds to circle inversion: en.wikipedia.org/wiki/Inversive_geometry – Qiaochu Yuan Mar 14 '11 at 10:13
The question was also asked (and answered) here: mathoverflow.net/questions/58405 – t.b. Mar 14 '11 at 12:20

The transformation $z\to\frac{1}{\bar{z}}$ is an inversion across the unit circle—the conjugation keeps the image at the same argument as the preimage. Inversion across the unit circle maps lines and circles to lines and circles, with only lines and circles that pass through $0$ (the center of the unit circle) mapping to lines.

$z\to\frac{1}{z}$ can be thought of as the inversion across the unit circle composed with a reflection over the real axis (complex conjugation $z\to\bar{z}$).

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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.

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+1 for the connection to complex analysis. A small note: Fractional linear transformations are also called Möbius transformations. – Srivatsan Dec 25 '11 at 23:47