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1992 Paper 2 Q10
D: 1600.0 B: 1529.8
complex numbers locus Argand diagram argument modulus circle inscribed angle theorem geometric proof cross-ratio

Let \(\alpha\) be a fixed angle, \(0 < \alpha \leqslant\frac{1}{2}\pi.\) In each of the following cases, sketch the locus of \(z\) in the Argand diagram (the complex plane):

  1. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha,}\)
  2. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha-\pi,}\)
  3. \(|\dfrac{z-1}{z}|=1.\)
Let \(z_{1},z_{2},z_{3}\) and \(z_{4}\) be four points lying (in that order) on a circle in the Argand diagram. If \[ w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \] show, by considering \(\arg w\), that \(w\) is real.

Solution:

TikZ diagram
TikZ diagram
TikZ diagram
TikZ diagram
\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real

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