Problems

Filters
Clear Filters

3 problems found

2017 Paper 3 Q2
D: 1700.0 B: 1500.1
complex numbers rotation transformation composition of transformations geometric proof exponential form commutativity

The transformation \(R\) in the complex plane is a rotation (anticlockwise) by an angle \(\theta\) about the point represented by the complex number \(a\). The transformation \(S\) in the complex plane is a rotation (anticlockwise) by an angle \(\phi\) about the point represented by the complex number \(b\).

  1. The point \(P\) is represented by the complex number~\(z\). Show that the image of \(P\) under \(R\) is represented by \[ \e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,. \]
  2. Show that the transformation \(SR\) (equivalent to \(R\) followed by \(S\)) is a rotation about the point represented by \(c\), where \[ %\textstyle c\,\sin \tfrac12 (\theta+\phi) = a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi \,, \] provided \(\theta+\phi \ne 2n\pi\) for any integer \(n\). What is the transformation \(SR\) if \(\theta +\phi = 2\pi\)?
  3. Under what circumstances is \(RS =SR\)?

Solution:

  1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
  2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
  3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.

View
2009 Paper 3 Q6
D: 1700.0 B: 1473.1
complex numbers modulus Ptolemy's theorem cyclic quadrilateral unit circle trigonometric identity geometric proof exponential form

Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.

View
1997 Paper 2 Q5
D: 1600.0 B: 1484.0
complex numbers exponential form locus curve sketching bijection polar coordinates circles rays

The complex numbers \(w=u+\mathrm{i}v\) and \(z=x+\mathrm{i}y\) are related by the equation $$z= (\cos v+\mathrm{i}\sin v)\mathrm{e}^u.$$ Find all \(w\) which correspond to \(z=\mathrm{i\,e}\). Find the loci in the \(x\)--\(y\) plane corresponding to the lines \(u=\) constant in the \(u\)--\(v\) plane. Find also the loci corresponding to the lines \(v=\) constant. Illustrate your answers with clearly labelled sketches. Identify two subsets \(W_1\) and \(W_2\) of the \(u\)--\(v\) plane each of which is in one-to-one correspondence with the first quadrant \(\{(x,\,y):\,x>0,\,y>0\}\) of the \(x\)--\(y\) plane. Identify also two subsets \(W_3\) and \(W_4\) each of which is in one-to-one correspondence with the set \(\{z\,:0<\,\vert z\vert\,<1\}\). \noindent[{\bf NB} `one-to-one' means here that to each value of \(w\) there is only one corresponding value of \(z\), and vice-versa.]

View