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CBSE · Class 11 · Mathematics

Complex Numbers and Quadratic Equations

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
i−33 equals:
  1. i
  2. 1
  3. −1
  4. −i
2.
(6 + 5i)(5 + 4i) equals:
  1. 50 + 49i
  2. 30 + 20i
  3. 10 + 49i
  4. 10 − 49i
3.
The modulus of −12 − 5i is:
  1. 7
  2. 17
  3. 169
  4. 13
4.
(1 − 3i) + (3 − i) equals:
  1. −2 − 4i
  2. 4 − 4i
  3. −2 − 2i
  4. 4 − 2i
5.
i2 equals:
  1. 0
  2. 1
  3. −1
  4. i
6.
The roots of x2 + 2x + 17 = 0 are:
  1. −1 + 4i and −1 − 4i
  2. 3 and −5
  3. −1 + 8i and −1 − 8i
  4. 1 + 4i and 1 − 4i
7.
The multiplicative inverse of 1 − 3i is:
  1. 1/10 − (3/10)i
  2. 1 + 3i
  3. 1/10 + (3/10)i
  4. −1/10 − (3/10)i
8.
Assertion (A): The modulus of 12 + 5i is 17.
Reason (R): |a + ib| = √(a2 + b2).
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Express (1 + 7i)(2 − 4i) in the form a + ib.
10.
Express (2 + 3i) ÷ (5 + 4i) in the form a + ib.
11.
Solve x2 + 3x + 4 = 0.
12.
If z = −2 + i, find z2 and z3.
13.
Simplify i38 + i47 + i11.
14.
Find real numbers x and y if (x + 6) + (y − 3)i = 4 + 6i.
15.
For z = 8 + 2i, find its conjugate, its modulus and its multiplicative inverse.
16.
For z1 = −20 + 21i and z2 = 4 + 3i, find |z1|, |z2| and |z1z2|. Check that |z1z2| = |z1| |z2|.
Section C — Word problems4 × 4 = 16 marks
17.
If z1 = 3 + 4i and z2 = 3 + 4i, find (a) z1 + z2, (b) z1 − z2, (c) z1z2, (d) z1 ÷ z2.
18.
Solve 2x2 − 4x + 10 = 0. Check that the sum of the roots is −b/a and their product is c/a.
19.
Show that (1 − i)2 = −2i, and hence find (1 − i)8.
20.
Find real numbers x and y such that (x + iy)2 = −2i, and so find the square roots of −2i.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Mathematics

Answer key

Complex Numbers and Quadratic Equations · AW-11M-CPX-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) −i−33 = 4 × −9 + 3
2(c) 10 + 49i30 + 49i + 20i2, and i2 = −1
3(d) 13√(122 + 52) = √169
4(b) 4 − 4icombine the real parts and the imaginary parts
5(c) −1—
6(a) −1 + 4i and −1 − 4iD = 4 − 68 = −64
7(c) 1/10 + (3/10)iz̄ ÷ |z|2 = (1 + 3i) ÷ 10
8(d) A is false, but R is true.A is false: √(122 + 52) = 13. R is true.
930 + 10i2 − 4i + 14i − 28i2
1022/41 + (7/41)imultiply top and bottom by 5 − 4i; the denominator becomes 41
11x = −3/2 ± (√7/2)iD = 9 − 16 = −7; x = (−3 ± √(−7)) ÷ 2
12z2 = 3 − 4i; z3 = −2 + 11iz3 = z2 × z
13−1 − 2ii38 = −1, i47 = −i, i11 = −i
14x = −2, y = 9equate the real parts and the imaginary parts
15z̄ = 8 − 2i; |z| = 2√17; z−1 = 2/17 − (1/34)i|z|2 = 68
16|z1| = 29, |z2| = 5, z1z2 = −143 + 24i, |z1z2| = 145 = 29 × 5—
17(a) 6 + 8i; (b) 0; (c) −7 + 24i; (d) 1for (d), multiply top and bottom by 3 − 4i
18x = 1 ± 2i, i.e. 1 + 2i and 1 − 2i. Sum = 2, and −b/a = 4/2 = 2; product = 5, and c/a = 10/2 = 5.D = −64
19(1 − i)2 = 1 − 2i + i2 = −2i; (1 − i)8 = (−2i)4 = 16(−1)4 × 24 × i4
20x2 − y2 = 0 and 2xy = −2, giving x = 1, y = −1 or x = −1, y = 1. Square roots: ±(1 − i)try x2 + y2 = |−2i| = 2