CBSE Important Questions Class 11 Maths Chapter 4 – Complex Numbers and Quadratic Equations

Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations is part of the CBSE Class 11 Maths syllabus. It introduces the imaginary unit i, where i² = −1, and explains how a complex number is written in the form a + ib. Students learn to add, subtract, multiply and divide complex numbers, find the conjugate and the modulus, write a number in polar form, and solve quadratic equations that have no real roots.

These CBSE important questions for Class 11 Maths Chapter 4 cover the powers of i, evaluating expressions with i, finding the modulus and conjugate, expressing a complex number in the form a + ib, solving quadratic equations with negative discriminant, and the condition for two quadratic equations to have a common root. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. If ax² + bx + c = 0 and bx² + cx + a = 0 have a common root, then find the value of (a³ + b³ + c³)/abc.
(a) 3
(b) 2
(c) 1
(d) None of these

Answer: (a) 3
When two such cyclic equations have a common root, the standard condition that comes out is a + b + c = 0.
Now we use the identity:
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Since a + b + c = 0, the whole right side becomes 0.
So a³ + b³ + c³ = 3abc
Therefore (a³ + b³ + c³)/abc = 3

Q2. Evaluate i³⁹ + i⁻⁶⁹.

Answer: We use the fact that i⁴ = 1, so we only need the remainder when the power is divided by 4.
For i³⁹: 39 = 4 × 9 + 3, so i³⁹ = i³ = −i
For i⁻⁶⁹: i⁻⁶⁹ = 1/i⁶⁹. Now 69 = 4 × 17 + 1, so i⁶⁹ = i.
So i⁻⁶⁹ = 1/i. Multiplying numerator and denominator by i:
= i/i² = i/(−1) = −i
Adding both: i³⁹ + i⁻⁶⁹ = −i + (−i) = −2i

Q3. What are the roots of the equation −2x² + 7x + 1 = 0?

Answer: Multiply throughout by −1 to make the leading coefficient positive:
2x² − 7x − 1 = 0
Here a = 2, b = −7, c = −1.
D = b² − 4ac = 49 + 8 = 57
Using the quadratic formula x = [−b ± √D] / 2a:
x = [7 ± √57] / 4
So the roots are x = (7 + √57)/4 and x = (7 − √57)/4.
Since D is positive, both roots are real and distinct.

Q4. Express (5i)(−3i/5) in the form a + ib.

Answer: Multiply the two terms:
(5i) × (−3i/5)
= (5 × −3/5) × (i × i)
= −3 × i²
Since i² = −1:
= −3 × (−1)
= 3
So in the form a + ib, the answer is 3 + 0i.

Q5. Find the modulus and the conjugate of the complex number z = 3 + 4i.

Answer: For z = a + ib:
Modulus |z| = √(a² + b²)
|z| = √(3² + 4²) = √(9 + 16) = √25 = 5
Conjugate z̄ = a − ib
z̄ = 3 − 4i
(Note also that z × z̄ = 9 + 16 = 25 = |z|², which is always true.)

Q6. Express (1 + i) / (1 − i) in the form a + ib.

Answer: Multiply the numerator and denominator by the conjugate of the denominator, which is (1 + i):
= [(1 + i)(1 + i)] / [(1 − i)(1 + i)]
Numerator: (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i
Denominator: (1)² − (i)² = 1 − (−1) = 2
So the value = 2i/2 = i
In the form a + ib, the answer is 0 + 1i.

Q7. Solve the quadratic equation x² + x + 1 = 0.

Answer: Here a = 1, b = 1, c = 1.
D = b² − 4ac = 1 − 4 = −3
Since D is negative, the roots are complex.
Using the quadratic formula:
x = [−1 ± √(−3)] / 2
Now √(−3) = √3 × √(−1) = √3 i
So x = (−1 ± √3 i) / 2
The roots are x = (−1 + √3 i)/2 and x = (−1 − √3 i)/2.

Q8. Find the multiplicative inverse of the complex number 4 − 3i.

Answer: The multiplicative inverse of z is 1/z, which equals z̄ / |z|².
Here z = 4 − 3i, so z̄ = 4 + 3i.
|z|² = 4² + (−3)² = 16 + 9 = 25
So the inverse = (4 + 3i) / 25
= 4/25 + (3/25)i
(Check: (4 − 3i) × (4 + 3i)/25 = 25/25 = 1 ✓)

Q9. If z = 2 − 3i, find z² and write it in the form a + ib.

Answer: z² = (2 − 3i)²
Using (a − b)² = a² − 2ab + b²:
= 4 − 12i + 9i²
Since i² = −1:
= 4 − 12i − 9
= −5 − 12i

Q10. Find the value of i¹⁰⁰ + i¹⁰¹ + i¹⁰² + i¹⁰³.

Answer: Divide each power by 4 and take the remainder, since i⁴ = 1.
100 ÷ 4 gives remainder 0, so i¹⁰⁰ = 1
101 gives remainder 1, so i¹⁰¹ = i
102 gives remainder 2, so i¹⁰² = −1
103 gives remainder 3, so i¹⁰³ = −i
Adding them: 1 + i − 1 − i = 0
(In fact the sum of any four consecutive powers of i is always 0.)

Q11. Express the complex number 1 + i√3 in polar form.

Answer: Let z = r(cosθ + i sinθ), where a = 1 and b = √3.
Modulus: r = √(a² + b²) = √(1 + 3) = 2
Argument: cosθ = a/r = 1/2 and sinθ = b/r = √3/2
Both are positive, so θ lies in the first quadrant and θ = π/3.
So the polar form is
z = 2(cos π/3 + i sin π/3)

Q12. What is the discriminant of a quadratic equation? State what it tells us about the nature of the roots.

Answer: For the quadratic equation ax² + bx + c = 0 (a ≠ 0), the discriminant is D = b² − 4ac.
• If D > 0, the roots are real and distinct.
• If D = 0, the roots are real and equal.
• If D < 0, the roots are complex conjugates of each other, of the form p + iq and p − iq.
In Class 11 we can now solve the third case too, which was not possible earlier.

Related Study Material on Extramarks

• CBSE Important Questions Class 11 Maths Chapter 1
• CBSE Important Questions Class 11 Maths Chapter 2
• CBSE Important Questions Class 11 Maths Chapter 3
• CBSE Important Questions Class 11 Maths Chapter 5
• CBSE Important Questions Class 11 Maths Chapter 6
• CBSE Important Questions Class 11 Maths Chapter 7
• CBSE Important Questions Class 11 Maths Chapter 8
• CBSE Important Questions Class 11 Maths Chapter 9
• CBSE Important Questions Class 11 Maths Chapter 10
• CBSE Important Questions Class 11 Maths Chapter 11
• CBSE Important Questions Class 11 Maths Chapter 12
• CBSE Important Questions Class 11 Maths Chapter 13
• CBSE Important Questions Class 11 Maths Chapter 14