1987 Paper 2 Q4

Year: 1987
Paper: 2
Question Number: 4

Course: UFM Pure
Section: Complex numbers 2

Difficulty: 1500.0 Banger: 1500.0

Problem

Explain the geometrical relationship between the points in the Argand diagram represented by the complex numbers \(z\) and \(z\mathrm{e}^{\mathrm{i}\theta}.\) Write down necessary and sufficient conditions that the distinct complex numbers \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle taken in anticlockwise order. Show that \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle (taken in any order) if and only if \[ \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta=0. \] Find necessary and sufficient conditions on the complex coefficients \(a,b\) and \(c\) for the roots of the equation \[ z^{3}+az^{2}+bz+c=0 \] to lie at the vertices of an equilateral triangle in the Argand digram.

Solution

The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)
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Problem source
Explain the geometrical relationship between the points in the Argand diagram represented by the complex numbers $z$ and $z\mathrm{e}^{\mathrm{i}\theta}.$ 
Write down necessary and sufficient conditions that the distinct complex numbers $\alpha,\beta$ and $\gamma$ represent the vertices of an equilateral triangle taken in anticlockwise order. 
Show that $\alpha,\beta$ and $\gamma$ represent the vertices of an equilateral triangle (taken in any order) if and only if 
\[
\alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta=0.
\]
Find necessary and sufficient conditions on the complex coefficients $a,b$ and $c$ for the roots of the equation 
\[
z^{3}+az^{2}+bz+c=0
\]
to lie at the vertices of an equilateral triangle in the Argand digram. 
Solution source
The point $ze^{i\theta}$ is obtained by rotating the point $z$ about $0$ by an angle $\theta$ anticlockwise.

The complex numbers $\alpha, \beta$ and $\gamma$ will form an equilateral triangle iff the angles between each side are $\frac{\pi}{3}$, ie

\begin{align*}
\begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\
{\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\
{\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases}
\end{align*}
We don't need all these equations, since the first two are equivalent to the third.

Combining the first two equations, we have

\begin{align*}
&& \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\
\Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\
\Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\
\Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0
\end{align*}
as required.

If the roots of $z^{3}+az^{2}+bz+c=0$ are $\alpha, \beta, \gamma$ then $\alpha+\beta+\gamma = -a$ and $\beta\gamma+\gamma\alpha+\alpha\beta = b$. We also have that $a^2 - 2b = \alpha^2+\beta^2+\gamma^2$. Therefore there roots will lie at the vertices of an equilateral triangle iff $a^2-3b = 0$