18 problems found
In this question, all gravitational forces are to be neglected. A rigid frame is constructed from 12 equal uniform rods, each of length \(a\) and mass \(m,\) forming the edges of a cube. Three of the edges are \(OA,OB\) and \(OC,\) and the vertices opposite \(O,A,B\) and \(C\) are \(O',A',B'\) and \(C'\) respectively. Forces act along the lines as follows, in the directions indicated by the order of the letters: \begin{alignat*}{3} 2mg\mbox{ along }OA, & \qquad & mg\mbox{ along }AC', & \qquad & \sqrt{2}mg\mbox{ along }O'A,\\ \sqrt{2}mg\mbox{ along }OA', & & 2mg\mbox{ along }C'B, & & mg\mbox{ along }A'C. \end{alignat*}
A path is made up in the Argand diagram of a series of straight line segments \(P_{1}P_{2},\) \(P_{2}P_{3},\) \(P_{3}P_{4},\ldots\) such that each segment is \(d\) times as long as the previous one, \((d\neq1)\), and the angle between one segment and the next is always \(\theta\) (where the segments are directed from \(P_{j}\) towards \(P_{j+1}\), and all angles are measured in the anticlockwise direction). If \(P_{j}\) represents the complex number \(z_{j},\) express \[ \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \] as a complex number (for each \(n\geqslant2\)), briefly justifying your answer. If \(z_{1}=0\) and \(z_{2}=1\), obtain an expression for \(z_{n+1}\) when \(n\geqslant2\). By considering its imaginary part, or otherwise, show that if \(\theta=\frac{1}{3}\pi\) and \(d=2\), then the path crosses the real axis infinitely often.
Solution: \begin{align*} && | \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} | &= d \\ && \arg \left ( \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \right) &= \arg (z_{n+1}-z_{n}) - \arg(z_{n}-z_{n-1}) \\ &&&= \theta \\ \Rightarrow && \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} &= d e^{i \theta} \end{align*} \begin{align*} && z_1 &= 0 \\ && z_2 &= 1 \\ && \frac{z_3-z_2}{z_2-z_1} &= de^{i \theta} \\ \Rightarrow && z_3 &= de^{i \theta} + 1 \\ && \frac{z_4-z_3}{z_3-z_2} &= de^{i \theta} \\ \Rightarrow && z_4 &= (d e^{i \theta})^2 + d e^{i \theta} + 1\\ \Rightarrow && z_{n+1} &= \frac{(de^{i \theta})^{n}-1}{de^{i \theta}-1} \end{align*} If \(d = 2, \theta = \tfrac13 \pi\), then, \(2e^{i \tfrac13 \pi} = 1 + \sqrt{3}i\) \begin{align*} \textrm{Im}(z_{n+1})) &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{2e^{i \tfrac13 \pi}-1}\right) \\ &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{\sqrt{3}i}\right) \\ &= -\frac{1}{\sqrt{3}}\textrm{Re} \left (2^n e^{i \frac{n}{3} \pi} \right) + \frac{1}{\sqrt{3}} \end{align*} Which clearly changes sign infinitely many times, ie crosses the origin infinitely many times.
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 \(a^2-3b = 0\)