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Complex Numbers: modulus-argument form, Argand geometry & a tangent identity — ASRJC 2025 H2 Maths Prelim Paper 2
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Question
The points \(P\) and \(Q\) are represented by the complex numbers \(p\) and \(q\) respectively, where \[ p = \sqrt{2} + \sqrt{2}\,\mathrm{i},\qquad \arg(q) = \frac{2\pi}{3},\qquad |q| = 2. \]
(a) Find \(|p|\) and \(\arg(p)\) in exact form.
(b) Sketch the points \(P\) and \(Q\) on an Argand diagram.
(c) Use the Argand diagram to deduce \(\mathrm{Re}(q)\) and \(\mathrm{Im}(q)\), giving your answers in exact form.
(d) The point \(R\) is represented by the complex number \(p + q\). What can you deduce about the shape of quadrilateral \(OPRQ\), where \(O\) is the origin?
(e) By considering \(\arg(p+q)\) or otherwise, show that \[ \tan\frac{11\pi}{24} = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2. \]
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(a) Modulus and argument of \(p\)
Use \(|p|=\sqrt{(\mathrm{Re})^2+(\mathrm{Im})^2}\) and the fact that \(p\) lies in the first quadrant.
\[ \begin{aligned} |p| &= \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2,\\ \arg(p) &= \tan^{-1}\!\left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{\pi}{4}. \end{aligned} \](b) Argand diagram
Plot \(P=(\sqrt{2},\sqrt{2})\), \(Q\) at modulus \(2\) and argument \(\frac{2\pi}{3}\), and \(R=p+q\) completing the parallelogram.
\(P=(\sqrt{2},\sqrt{2})\approx(1.41,1.41)\); \(Q=(-1,\sqrt{3})\approx(-1,1.73)\); \(R=p+q=(\sqrt{2}-1,\;\sqrt{2}+\sqrt{3})\approx(0.41,3.15)\). The points \(O\), \(P\), \(R\), \(Q\) form a rhombus with \(OP\) and \(OQ\) as adjacent sides.
(c) Real and imaginary parts of \(q\)
Drop a perpendicular from \(Q\) to the imaginary axis and read off the horizontal and vertical components.
Let \(F\) be the foot of the perpendicular from \(Q\) to the \(y\)-axis.
\[ \begin{aligned} \operatorname{Re}(q) = -QF &= -|q|\sin\frac{2\pi}{3} = -1, \\ \operatorname{Im}(q) = OF &= |q|\cos\frac{2\pi}{3} = \sqrt{3}. \end{aligned} \]So \(\mathrm{Re}(q) = -1\) and \(\mathrm{Im}(q) = \sqrt{3}\).
(d) Shape of \(OPRQ\)
Equal adjacent sides of a parallelogram force a rhombus.
Since \(|p| = |q| = 2\), \(OPRQ\) is a parallelogram with equal adjacent sides, so quadrilateral \(OPRQ\) is a rhombus.
(e) Deriving \(\tan\frac{11\pi}{24}\)
In a rhombus the diagonal \(OR\) bisects \(\angle POQ\), giving \(\arg(p+q)\) as the average of \(\arg(p)\) and \(\arg(q)\); then take the tangent of the explicit \(p+q\).
\[ \begin{aligned} \arg(p+q) &= \frac{1}{2}\bigl(\arg(p) + \arg(q)\bigr)\\ & = \frac{1}{2}\left(\frac{\pi}{4} + \frac{2\pi}{3}\right)\\ & = \frac{1}{2}\cdot\frac{11\pi}{12}\\ & = \frac{11\pi}{24}. \end{aligned} \]Also, \(p + q = (\sqrt{2}+\sqrt{2}\,\mathrm{i}) + (-1+\sqrt{3}\,\mathrm{i}) = (\sqrt{2}-1) + (\sqrt{2}+\sqrt{3})\,\mathrm{i}.\)
Therefore \[ \tan\frac{11\pi}{24} = \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-1}. \]
Rationalise by multiplying numerator and denominator by \((\sqrt{2}+1)\):
\[ \begin{aligned} \tan\frac{11\pi}{24} &= \frac{(\sqrt{2}+\sqrt{3})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\\ &= \frac{2 + \sqrt{2} + \sqrt{6} + \sqrt{3}}{2 - 1}\\ &= \sqrt{6} + \sqrt{3} + \sqrt{2} + 2. \qquad \text{(Shown)} \end{aligned} \]