Complex Numbers Question (1 Viewer)

altSwift · Jan 24, 2018

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

InteGrand · Jan 25, 2018

altSwift said:
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!!).)$

$\noindent Good luck!$

Complex Numbers Question (1 Viewer)

altSwift

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InteGrand

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