Complex Numbers: Argand geometry, rhombus & reflections — NYJC 2025 H2 Maths Prelim Paper 1
What this question tests
Question
The points \(P\), \(Q\) and \(R\) representing the complex numbers \(p\), \(q\) and \(r\) on an Argand diagram are such that \(\arg(p) = \alpha\) and \(\arg(q) = \beta\), where \(0 < \alpha < \beta < \dfrac{\pi}{2}\), \(\beta > 2\alpha\) and \(r = p + q\).
(a) If \(|p| = |q|\), describe the shape of the quadrilateral \(OPRQ\). Hence find \(\arg(r)\) in terms of \(\alpha\) and \(\beta\). [3]
(b) The point \(Q'\), representing the complex number \(q'\), is the reflection of the point \(Q\) in \(OP\). State the angle \(POQ'\). [1]
By leaving your answers in terms of \(\alpha\), \(\beta\) and \(|q|\) where applicable, hence, or otherwise,
(i) find the argument of the complex number \(q'\), [1]
(ii) find the real and imaginary parts of \(q'\) and write down \(q'\) in \(a + \mathrm{i}b\) form. [3]
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(a) Shape of \(OPRQ\) and \(\arg(r)\)
Equal moduli make the addition parallelogram a rhombus, whose diagonal \(OR\) bisects \(\angle POQ\).
If \(|p|=|q|\), then \(OPRQ\) forms a rhombus.
The diagonal \(OR\) bisects the angle between \(OP\) and \(OQ\), so
\[ \arg(r) = \tfrac{1}{2}(\beta-\alpha)+\alpha = \tfrac{1}{2}(\alpha+\beta) \](b) Angle \(POQ'\)
Reflecting \(Q\) in line \(OP\) preserves the angle to \(OP\), placing \(Q'\) as far below \(OP\) as \(Q\) is above it.
\(\angle POQ' = \beta - \alpha\)
(b)(i) Argument of \(q'\)
Subtract the reflection angle from \(\arg(p)\); since \(\beta>2\alpha\) the result is negative, so \(q'\) lies below the real axis.
Method 1: \(\arg(q') = -(\beta - 2\alpha) = 2\alpha - \beta\) (since \(\beta > 2\alpha\), \(q'\) is below the real axis).
Method 2: \(\arg(q') = \alpha - (\beta-\alpha) = 2\alpha - \beta\).
(b)(ii) Real and imaginary parts of \(q'\)
With \(|q'|=|q|\), drop a perpendicular from \(Q'\) to the real axis and read off the adjacent and opposite sides.
\(|q'| = |q|\). Let \(F\) be the foot of perpendicular from \(Q'\) to the real axis.
Real part of \(q' = |q|\cos(-\beta+2\alpha)\) (adjacent side of \(\triangle OQ'F\)).
Imaginary part of \(q' = |q|\sin(-\beta+2\alpha)\) (opposite side of \(\triangle OQ'F\), and negative).
\[ \therefore\; q' = |q|\cos(\beta-2\alpha) - \mathrm{i}|q|\sin(\beta-2\alpha) \]