1. ## Complex Numbers Question

z1, z2 and z3 lie on a circle. The origin also lies on the circle
Prove 1/z1, 1/z2 and 1/z3 are collinear

2. ## Re: Complex Numbers Question

Originally Posted by altSwift
z1, z2 and z3 lie on a circle. The origin also lies on the circle
Prove 1/z1, 1/z2 and 1/z3 are collinear
$\noindent Here's some steps/hints for a possible method:$

$\noindent 0. Show in general that if u,v,w\in \mathbb{C} are collinear, then A u, A v, A w are collinear for any given complex number A.$

$\noindent 1. Show that the result is true if the circle is centred at the point (R,0) at the x-axis, where R>0 is the circle's radius. (Hint to show this: on this circle, a point z can be written as z = R + R\,\mathrm{cis}(\theta). Show with the help of trig. identities that this implies that \mathrm{Re}\left(\frac{1}{z}\right) = \frac{1}{2R}, which is a constant (doesn't depend on \theta). This implies the three points z_1,z_2, z_3 will all lie on the same line, namely the vertical line \mathrm{Re}(z) = \frac{1}{2R}.)$

$\noindent 2. Show that by proving 1. above, this is enough to deduce the result for \emph{any} circle described in the question, not necessarily centred on the x-axis. (Hint for this: given an arbitrary circle described by the question, we can rotate the circle onto the x-axis by multiplying by \mathrm{cis}(\alpha) for some appropriate angle \alpha. So if the points z_k lie on some circle described by the question, then the points z_k \mathrm{cis}(\alpha) lie on a circle like that in 1. Hence by what is proved in 1., the points \frac{1}{z_k \mathrm{cis}(\alpha)} are collinear. Now explain why this implies the z_k are collinear (refer to step 0!!).)$

$\noindent Good luck!$

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