The points $z_1, z_2, z_3$ are three complex numbers lying on a the circumference of a circle passing through the origin in the Argand diagram. Show that $\frac 1{z_1} , \frac 1{z_2}$ and $\frac 1{z_3}$, in the Argand diagram, are collinear

I got a cartesian equation of the circle, so $(x-h)^2 + (y-k)^2 = h^2 + k^2$, but then I seem to be introducing too many new variables which indicates I'm not approaching this the best way.

  • $\begingroup$ if a line passes through a circle it can touch it only at two points so either two points must be the same if they are collinear. $\endgroup$ – Jasser Sep 21 '14 at 12:07

It suffices to show that any under the map $z \mapsto \frac{1}{z}$ any circle passing through the origin is mapped to a line. This follows from some basic geometric facts about the complex plane, but typically those aren't available when someone would encounter such a problem.

One circle passing through the origin the (radius $\tfrac{1}{2}$) circle parameterized by the polar equation $r_0(\theta) := \cos \theta$, and any other circle is given by dilating and rotating this circle about the origin, and so is given by $$r(\theta) := A r_0(\theta - \theta_0) = A \cos (\theta - \theta_0).$$

Under the map $i: z \mapsto \frac{1}{z}$, this curve is mapped to the curve parameterized by $$i(r(\theta)) = \frac{1}{A} \sec(\theta - \theta_0).$$ But this is precisely the polar parameterization of the line intersecting and perpendicular to the ray that (1) makes a (signed) angle $\theta_0$ with the $x$-axis and whose distance to the origin is $\frac{1}{A}$.


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