Complex Numbers: Argand diagram geometry, multiplication as rotation, and a trig identity — VJC 2025 H2 Math Prelim P2
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Question
Do not use a calculator in answering this question.
The complex number \(z\) has modulus \(1\) and argument \(\theta\), where \(\dfrac{\pi}{2} < \theta < \pi\), and the complex number \(w\) is given by \(w = \mathrm{i}\sqrt{3}\,z\). The point \(P\) on the Argand diagram represents \(z\) and lies in the second quadrant.
(a) On the copy of the Argand diagram with origin \(O\) in the Printed Answer Booklet, plot the points \(Q\) and \(R\) to represent \(w\) and \(z - w\) respectively. Show clearly the geometrical relationship between the points \(P\), \(Q\) and \(R\).
(b) Find the area of the quadrilateral \(ORPQ\).
It is given that \(\theta = \dfrac{3\pi}{4}\).
(c) Find \(z\) in the form \(x + y\mathrm{i}\), where \(x\) and \(y\) are real numbers.
(d) Show that \(z - w = k\!\left[\left(\sqrt{3}-1\right) + \left(\sqrt{3}+1\right)\mathrm{i}\right]\), where \(k\) is a constant to be determined.
(e) Hence show that \(\tan\dfrac{5\pi}{12} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}\).
Show full worked solution▾
(a) Plotting \(Q\) and \(R\)
Multiplying by \(\mathrm{i}\sqrt{3}\) scales \(z\) by \(\sqrt{3}\) and rotates it by \(\dfrac{\pi}{2}\); the relationships then follow from the moduli and arguments.
Since \(w = \mathrm{i}\sqrt{3}\,z\), we have \(|w| = \sqrt{3}\) and \(\arg w = \arg z + \dfrac{\pi}{2}\), so \(OQ \perp OP\) at \(O\) with \(|OP| = 1\) and \(|OQ| = \sqrt{3}\). Also \(z - w = z(1 - \mathrm{i}\sqrt{3})\) with \(|1 - \mathrm{i}\sqrt{3}| = 2\), so \(|OR| = 2\); and \(\overrightarrow{PR} = -\overrightarrow{OQ}\), so \(PR \perp OP\) at \(P\) with \(|PR| = \sqrt{3}\). Hence \(R\) and \(Q\) lie on opposite sides of line \(OP\), each at perpendicular distance \(\sqrt{3}\) from \(OP\).
(b) Area of \(ORPQ\)
Treat \(OP\) as a diagonal of length \(1\); the two triangles on either side each have height \(\sqrt{3}\).
\[ \text{Area}(ORPQ) = \dfrac{1}{2}(1)(\sqrt{3} + \sqrt{3}) = \sqrt{3}. \](c) \(z\) in Cartesian form
With \(\theta = \dfrac{3\pi}{4}\), drop a perpendicular from \(P\) to the real axis and use the reference angle \(\dfrac{\pi}{4}\).
\[ \begin{aligned} \text{Length } x &= 1\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \\ \text{Length } y &= 1\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \end{aligned} \] \[ \therefore z = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\,\mathrm{i} \](d) Showing the form of \(z - w\)
Substitute the Cartesian \(z\) into \(z(1 - \mathrm{i}\sqrt{3})\) and expand.
\[ \begin{aligned} z - w &= z(1 - \mathrm{i}\sqrt{3}) = \left(-\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\mathrm{i}\right)(1 - \mathrm{i}\sqrt{3})\\ &= \dfrac{1}{\sqrt{2}}(-1 + \mathrm{i})(1 - \mathrm{i}\sqrt{3})\\ &= \dfrac{1}{\sqrt{2}}\!\left(-1 + \mathrm{i}\sqrt{3} + \mathrm{i} + \sqrt{3}\right)\\ &= \dfrac{1}{\sqrt{2}}\!\left[(\sqrt{3} - 1) + (\sqrt{3} + 1)\mathrm{i}\right] \end{aligned} \]Hence \(k = \dfrac{1}{\sqrt{2}}\).
(e) Deducing \(\tan\dfrac{5\pi}{12}\)
The argument of \(z - w\) equals \(\dfrac{3\pi}{4} - \dfrac{\pi}{3} = \dfrac{5\pi}{12}\), so \(\tan\) of that argument is its imaginary part over its real part.
