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Complex Numbers and Quadratic Equations – Class 11 Mathematics notes, important questions and MCQs

NCERT Class 11 Mathematics, chapter 4. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Complex numbers a + ib extend real numbers so that every quadratic equation has roots. They can be added, multiplied, conjugated and shown on the Argand plane.

Algebra

  • i = √−1, i² = −1, i³ = −i, i⁴ = 1.
  • Conjugate z̄ = a − ib; modulus |z| = √(a² + b²); z·z̄ = |z|².
  • 1/z = z̄/|z|².

Polar form and quadratics

  • z = r(cos θ + i sin θ), r = |z|, θ = arg z.
  • Roots of ax² + bx + c = 0: x = [−b ± √(b² − 4ac)]/2a; if D < 0, roots are complex conjugates.
  • Sum of roots = −b/a, product = c/a.

Key terms

Modulus
distance of z from the origin, √(a² + b²)
Conjugate
a − ib for z = a + ib
Argument
angle θ the line joining z to the origin makes with the positive real axis; principal value in (−π, π]

Formulas

  • |z| = √(a² + b²)
  • x = [−b ± √(b² − 4ac)]/2a

Important questions with answers

Express (5 + 2i)/(1 − i) in a + ib form.2 marks

Multiply by (1 + i): (5 + 5i + 2i + 2i²)/2 = (3 + 7i)/2 = 3/2 + (7/2)i.

Solve x² + 3x + 9 = 0.3 marks

D = 9 − 36 = −27. x = (−3 ± √−27)/2 = (−3 ± 3√3 i)/2.

Convert z = −1 + i√3 to polar form, and find the modulus and argument of z².5 marks

r = √(1 + 3) = 2. The point is in Quadrant II: tan α = √3, α = π/3, so θ = π − π/3 = 2π/3. z = 2(cos 2π/3 + i sin 2π/3). z² has modulus r² = 4 and argument 4π/3, which in principal form is 4π/3 − 2π = −2π/3; z² = −2 − 2√3 i.

Quick revision

  • i² = −1.
  • Complex roots occur in conjugate pairs (real coefficients).
  • z·z̄ = |z|².

Exam tips

  • iⁿ repeats every 4: find n mod 4.
  • To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator.

Practice MCQs

  1. i² equals

    1. −1
    2. 1
    3. i
    4. −i
    Show answer

    Answer: (a) −1. Definition.

  2. Modulus of 3 + 4i is

    1. 5
    2. 7
    3. 25
    4. 1
    Show answer

    Answer: (a) 5. √(9 + 16).

  3. i¹⁰⁵ equals

    1. i
    2. 1
    3. −1
    4. −i
    Show answer

    Answer: (a) i. 105 mod 4 = 1.

  4. Conjugate of 2 − 5i is

    1. 2 + 5i
    2. −2 + 5i
    3. −2 − 5i
    4. 5 − 2i
    Show answer

    Answer: (a) 2 + 5i. Change sign of imaginary part.

  5. (2 + 3i)(1 − i) equals

    1. 5 + i
    2. −1 + 5i
    3. 5 − i
    4. 2 − 3i
    Show answer

    Answer: (a) 5 + i. 2 − 2i + 3i + 3.

  6. Roots of x² + x + 1 = 0 are

    1. real and equal
    2. real and distinct
    3. complex conjugates
    4. zero
    Show answer

    Answer: (c) complex conjugates. D = −3.

  7. (1 + i)² equals

    1. 2i
    2. 2
    3. 0
    4. 1 + 2i
    Show answer

    Answer: (a) 2i. 1 + 2i − 1.

  8. The multiplicative inverse of 3 + 4i is

    1. (3 − 4i)/25
    2. (3 + 4i)/25
    3. 3 − 4i
    4. (3 − 4i)/5
    Show answer

    Answer: (a) (3 − 4i)/25. z̄/|z|².

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.