Maths · Algebra · chapter 4 of 14

Class 11 Maths Chapter 4: Complex Numbers and Quadratic Equations case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

A school robotics club in Nagpur marks positions on its practice floor as complex numbers: the real axis points east, the imaginary axis points north, the origin O is the charging dock, and one unit is one metre. A rover starts at A = 2 + 3i. The command TURN multiplies the rover's present position by i. The command MIRROR replaces its present position by the conjugate. Riya wants to know where the rover ends up after each command, and how far it travels in a straight line.

Part 1 · 1 mark

After one MIRROR from A, the rover is at

(a) −2 + 3i(b) 2 − 3i(c) −2 − 3i(d) 3 + 2i

Show answer

Answer: (b) 2 − 3i

The conjugate of 2 + 3i is 2 − 3i, the mirror image of (2, 3) in the real axis: the point (2, −3).

Part 2 · 1 mark

After one TURN from A, the rover is at

(a) 3 − 2i(b) −2 + 3i(c) 2 + 3i(d) −3 + 2i

Show answer

Answer: (d) −3 + 2i

i(2 + 3i) = 2i + 3i² = −3 + 2i, the point (−3, 2).

Part 3 · 2 marks

Find the straight-line distance from A to the rover's position after one TURN from A. Show also that neither command changes the rover's distance from the dock O.

Answer and marking scheme

After TURN the rover is at −3 + 2i, so the distance is |(−3 + 2i) − (2 + 3i)| = |−5 − i| = √26 m. For any position z, |iz| = |i||z| = |z| and |z̄| = |z|, so both commands keep the distance √13 m from O.

|(−3 + 2i) − (2 + 3i)| = |−5 − i| = √26 m (1)
|iz| = |i||z| = |z| and |z̄| = |z| (1)

The rover starts at A and is given TURN twice. Find its final position, and show that it is the point opposite A through the dock O. How far is it from A?

Answer and marking scheme

Two TURNs multiply by i² = −1: the final position is −(2 + 3i) = −2 − 3i, the point (−2, −3), which is A reflected through O. Its distance from A is |(−2 − 3i) − (2 + 3i)| = |−4 − 6i| = √52 = 2√13 m, twice |A| = √13.

final position −2 − 3i = −A, opposite A through O (1)
distance |−4 − 6i| = 2√13 m (1)

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