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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.
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
i² equals
- −1
- 1
- i
- −i
Show answer
Answer: (a) −1. Definition.
Modulus of 3 + 4i is
- 5
- 7
- 25
- 1
Show answer
Answer: (a) 5. √(9 + 16).
i¹⁰⁵ equals
- i
- 1
- −1
- −i
Show answer
Answer: (a) i. 105 mod 4 = 1.
Conjugate of 2 − 5i is
- 2 + 5i
- −2 + 5i
- −2 − 5i
- 5 − 2i
Show answer
Answer: (a) 2 + 5i. Change sign of imaginary part.
(2 + 3i)(1 − i) equals
- 5 + i
- −1 + 5i
- 5 − i
- 2 − 3i
Show answer
Answer: (a) 5 + i. 2 − 2i + 3i + 3.
Roots of x² + x + 1 = 0 are
- real and equal
- real and distinct
- complex conjugates
- zero
Show answer
Answer: (c) complex conjugates. D = −3.
(1 + i)² equals
- 2i
- 2
- 0
- 1 + 2i
Show answer
Answer: (a) 2i. 1 + 2i − 1.
The multiplicative inverse of 3 + 4i is
- (3 − 4i)/25
- (3 + 4i)/25
- 3 − 4i
- (3 − 4i)/5
Show answer
Answer: (a) (3 − 4i)/25. z̄/|z|².
All chapters of Class 11 Mathematics
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.