Complex Numbers: Argand geometry, rhombus & reflections — NYJC 2025 H2 Math Prelim P1
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).
\[ \therefore\; q' = |q|\cos(\beta-2\alpha) - \mathrm{i}|q|\sin(\beta-2\alpha) \]