Complex Numbers: rotation of complex numbers & parallelogram geometry — RVHS 2025 H2 Math Prelim P2
River Valley High School2025 PrelimPaper 2●●○ Standard5 marks
What this question tests
Using multiplication by \(-\mathrm{i}\) to rotate a vector \(90^\circ\) clockwise in the Argand plane, and the midpoint-of-diagonals property of a parallelogram to locate a fourth vertex.
Question
The points \(A\), \(B\) and \(C\) represent the complex numbers \(z_A = 5+6\mathrm{i}\), \(z_B = 9+3\mathrm{i}\) and \(z_C\) respectively. \(ABC\) is an isosceles triangle labelled in a clockwise direction where \(\angle CAB = 90^\circ\).
(a) Find \(z_C\).
(b) The point \(D\) representing the complex number \(z_D\), is such that \(ABDC\) is a parallelogram. Find \(z_D\).
Show full worked solution▾
(a) Find \(z_C\)
The segment \(AC\) is obtained by rotating \(AB\) by \(90^\circ\) clockwise about \(A\), which corresponds to multiplying the vector \(z_B-z_A\) by \(-\mathrm{i}\).
\[ z_C - z_A = -\mathrm{i}(z_B - z_A) \] \[ z_C = z_A - \mathrm{i}(z_B - z_A) \] \[ z_C = (5+6\mathrm{i}) - \mathrm{i}\bigl[(9+3\mathrm{i}) - (5+6\mathrm{i})\bigr] = (5+6\mathrm{i}) - \mathrm{i}[4-3\mathrm{i}] = 2 + 2\mathrm{i} \](b) Find \(z_D\)
In parallelogram \(ABDC\) the diagonals \(AD\) and \(BC\) bisect each other, so their midpoints coincide.
Method — midpoints of diagonals of parallelogram \(ABDC\)
\[ \frac{z_A + z_D}{2} = \frac{z_B + z_C}{2} \]
\[ (5+6\mathrm{i}) + z_D = (9+3\mathrm{i}) + (2+2\mathrm{i}) \]
\[ z_D = 11 + 5\mathrm{i} - (5+6\mathrm{i}) = 6 - \mathrm{i} \]Answers: (a) \(z_C = 2+2\mathrm{i}\); (b) \(z_D = 6-\mathrm{i}\)