Important Questions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations 2026–27

Complex Numbers and Quadratic Equations introduces students to i, complex numbers, modulus, conjugate, Argand plane and complex roots. For CBSE Class 11 Maths, Chapter 4 helps students solve quadratic equations that do not have real roots.

Class 11 Maths Chapter 4 explains why the real number system is extended to complex numbers. Students learn that i² = −1 and that a number of the form a + ib is called a complex number. The chapter also covers real part and imaginary part, powers of i, algebra of complex numbers, modulus, conjugate, multiplicative inverse, Argand plane and polar representation.

Use these Important Questions Class 11 Maths Chapter 4 to practise NCERT-based questions for the 2026–27 exams. Begin with definitions and powers of i, then move to operations, conjugates, modulus and roots of quadratic equations.

Key Takeaways

• Chapter Name: Chapter 4 is Complex Numbers and Quadratic Equations in the 2026–27 NCERT Class 11 Maths book.
• Core Concept: A complex number is written as a + ib, where a and b are real numbers and i² = −1.
• Important Topics: Powers of i, modulus, conjugate, multiplicative inverse, Argand plane and polar form are key areas.
• Exam Focus: Students should practise simplification, standard form, complex roots and proof-based questions.

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Important Formulas for CBSE Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Before solving Important Questions for Class 11 Maths Chapter 4, revise these formulas. They help in simplifying powers of i, writing expressions in a + ib form and solving quadratic equations with complex roots.

Very Short Answer Questions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

These Class 11 Maths Chapter 4 questions test definitions, powers of i, real part, imaginary part and basic calculations.

Q1. What is the value of i²?

i² = −1

This is the basic property of iota.

Q2. Express √−25 in terms of i.

√−25 = √25 × √−1

= 5i

So, √−25 = 5i.

Q3. Write the real part and imaginary part of z = 7 − 4i.

For z = 7 − 4i,

Real part = 7

Imaginary part = −4

Q4. Find the value of i⁴.

i² = −1

i⁴ = (i²)² = (−1)² = 1

So, i⁴ = 1.

Q5. Find the value of i¹⁰.

i¹⁰ = i⁸ × i²

= (i⁴)² × i²

= 1² × (−1)

= −1

So, i¹⁰ = −1.

Q6. Find the conjugate of 6 + 5i.

The conjugate of a + ib is a − ib.

So, the conjugate of 6 + 5i is 6 − 5i.

Q7. Find the modulus of 3 + 4i.

For z = 3 + 4i,

|z| = √(3² + 4²)

= √(9 + 16)

= √25

= 5

Q8. What is the additive inverse of −2 + 7i?

The additive inverse of z is −z.

So, the additive inverse of −2 + 7i is 2 − 7i.

Q9. What is the multiplicative identity for complex numbers?

The multiplicative identity is 1 + 0i.

It is usually written as 1.

Q10. In which plane are complex numbers represented geometrically?

Complex numbers are represented in the Argand plane.

The horizontal axis is the real axis and the vertical axis is the imaginary axis.

Short Answer Questions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

These Complex Numbers and Quadratic Equations Class 11 important questions need 2 to 3 steps. Write the answer in standard form wherever required.

Q11. Express (3 + 2i) + (4 − 5i) in the form a + ib.

Add the real parts and imaginary parts separately.

(3 + 2i) + (4 − 5i)

= (3 + 4) + (2i − 5i)

= 7 − 3i

So, the answer is 7 − 3i.

Q12. Express (8 − 3i) − (2 + 5i) in the form a + ib.

Subtract real and imaginary parts.

(8 − 3i) − (2 + 5i)

= 8 − 3i − 2 − 5i

= 6 − 8i

So, the answer is 6 − 8i.

Q13. Express (2 + 3i)(4 − i) in the form a + ib.

Multiply the terms.

(2 + 3i)(4 − i)

= 8 − 2i + 12i − 3i²

= 8 + 10i − 3(−1)

= 11 + 10i

So, the answer is 11 + 10i.

Q14. Express (1 − i)² in the form a + ib.

(1 − i)² = 1² − 2i + i²

= 1 − 2i − 1

= −2i

So, the answer is 0 − 2i.

Q15. Find the value of i⁻³⁹.

Since powers of i repeat after every 4 powers,

i⁻³⁹ = 1 / i³⁹

Now, 39 divided by 4 leaves remainder 3.

So, i³⁹ = i³ = −i

Therefore,

i⁻³⁹ = 1 / (−i)

= −1/i

Multiply numerator and denominator by i.

= −i / i²

= −i / (−1)

= i

So, i⁻³⁹ = i.

Q16. Find the multiplicative inverse of 2 − 3i.

Let z = 2 − 3i.

The conjugate of z is 2 + 3i.

|z|² = 2² + (−3)² = 4 + 9 = 13

So,

1/z = z̄ / |z|²

= (2 + 3i) / 13

= 2/13 + 3i/13

Hence, the multiplicative inverse of 2 − 3i is 2/13 + 3i/13.

Q17. Express (5 + 2i) / (1 − 2i) in the form a + ib.

Multiply numerator and denominator by the conjugate of 1 − 2i, which is 1 + 2i.

(5 + 2i) / (1 − 2i)

= [(5 + 2i)(1 + 2i)] / [(1 − 2i)(1 + 2i)]

Numerator:

(5 + 2i)(1 + 2i) = 5 + 10i + 2i + 4i²

= 5 + 12i − 4

= 1 + 12i

Denominator:

(1 − 2i)(1 + 2i) = 1 − (2i)²

= 1 − 4i²

= 1 + 4

= 5

So,

(5 + 2i) / (1 − 2i) = (1 + 12i) / 5

= 1/5 + 12i/5

Q18. If 4x + i(3x − y) = 3 − 6i, find x and y.

Equate real and imaginary parts.

Real parts:

4x = 3

x = 3/4

Imaginary parts:

3x − y = −6

Substitute x = 3/4.

3(3/4) − y = −6

9/4 − y = −6

−y = −6 − 9/4

−y = −33/4

y = 33/4

So, x = 3/4 and y = 33/4.

Q19. Find the modulus and conjugate of −5 − 12i.

For z = −5 − 12i,

Modulus:

|z| = √[(-5)² + (-12)²]

= √(25 + 144)

= √169

= 13

Conjugate:

z̄ = −5 + 12i

So, the modulus is 13 and the conjugate is −5 + 12i.

Q20. Solve x² + 4 = 0.

x² + 4 = 0

x² = −4

x = ±√−4

x = ±2i

So, the roots are 2i and −2i.

Long Answer Questions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

These Class 11 Maths Complex Numbers Important Questions need step-wise solutions. Use properties of i, conjugate, modulus and quadratic equations carefully.

Q21. Express (5 − 3i)³ in the form a + ib.

Use (a − b)³ = a³ − 3a²b + 3ab² − b³.

Here, a = 5 and b = 3i.

(5 − 3i)³

= 5³ − 3(5²)(3i) + 3(5)(3i)² − (3i)³

= 125 − 225i + 15(9i²) − 27i³

Now, i² = −1 and i³ = −i.

= 125 − 225i + 135(−1) − 27(−i)

= 125 − 225i − 135 + 27i

= −10 − 198i

So, (5 − 3i)³ = −10 − 198i.

Q22. Find the value of (1 − i)⁴.

First find (1 − i)².

(1 − i)² = 1 − 2i + i²

= 1 − 2i − 1

= −2i

Now,

(1 − i)⁴ = [(1 − i)²]²

= (−2i)²

= 4i²

= −4

So, (1 − i)⁴ = −4.

Q23. Prove that z z̄ = |z|² for z = a + ib.

Let z = a + ib.

Then, z̄ = a − ib.

Now,

z z̄ = (a + ib)(a − ib)

= a² − iab + iab − i²b²

= a² − i²b²

Since i² = −1,

= a² + b²

Also,

|z| = √(a² + b²)

So,

|z|² = a² + b²

Hence,

z z̄ = |z|²

Q24. If x + iy = (a + ib) / (a − ib), prove that x² + y² = 1.

Given,

x + iy = (a + ib) / (a − ib)

Multiply numerator and denominator by the conjugate of denominator, a + ib.

x + iy = [(a + ib)(a + ib)] / [(a − ib)(a + ib)]

= (a + ib)² / (a² + b²)

= (a² + 2abi + i²b²) / (a² + b²)

= (a² − b² + 2abi) / (a² + b²)

So,

x = (a² − b²) / (a² + b²)

and

y = 2ab / (a² + b²)

Now,

x² + y²

= [(a² − b²)² / (a² + b²)²] + [4a²b² / (a² + b²)²]

= [(a² − b²)² + 4a²b²] / (a² + b²)²

= [a⁴ − 2a²b² + b⁴ + 4a²b²] / (a² + b²)²

= [a⁴ + 2a²b² + b⁴] / (a² + b²)²

= (a² + b²)² / (a² + b²)²

= 1

Hence proved.

Q25. Solve the quadratic equation x² + 2x + 5 = 0.

Compare x² + 2x + 5 = 0 with ax² + bx + c = 0.

Here,

a = 1, b = 2, c = 5

Discriminant:

D = b² − 4ac

= 2² − 4(1)(5)

= 4 − 20

= −16

Since D < 0, the roots are complex.

Using the quadratic formula,

x = [−b ± √D] / 2a

= [−2 ± √−16] / 2

= [−2 ± 4i] / 2

= −1 ± 2i

So, the roots are −1 + 2i and −1 − 2i.

Q26. Solve the quadratic equation 2x² + 3x + 5 = 0.

Here,

a = 2, b = 3, c = 5

D = b² − 4ac

= 3² − 4(2)(5)

= 9 − 40

= −31

Using the quadratic formula,

x = [−b ± √D] / 2a

= [−3 ± √−31] / 4

= [−3 ± i√31] / 4

So, the roots are:

x = (−3 + i√31) / 4 and x = (−3 − i√31) / 4

Q27. If one root of a quadratic equation with real coefficients is 3 + 2i, what is the other root?

For a quadratic equation with real coefficients, complex roots occur in conjugate pairs.

So, if one root is 3 + 2i, the other root is 3 − 2i.

Q28. Form a quadratic equation whose roots are 2 + 3i and 2 − 3i.

Let the roots be α = 2 + 3i and β = 2 − 3i.

Sum of roots:

α + β = (2 + 3i) + (2 − 3i) = 4

Product of roots:

αβ = (2 + 3i)(2 − 3i)

= 2² + 3²

= 4 + 9

= 13

The quadratic equation is:

x² − (sum of roots)x + product of roots = 0

x² − 4x + 13 = 0

Q29. Find the modulus of [(1 + i) / (1 − i)].

Let z = (1 + i) / (1 − i).

Multiply numerator and denominator by 1 + i.

z = [(1 + i)(1 + i)] / [(1 − i)(1 + i)]

= (1 + 2i + i²) / (1 − i²)

= (1 + 2i − 1) / (1 + 1)

= 2i / 2

= i

So,

|z| = |i| = 1

Q30. Represent z = 1 + i in polar form.

For z = 1 + i,

x = 1 and y = 1

Modulus:

r = √(1² + 1²)

= √2

Argument:

tan θ = y/x = 1/1 = 1

So, θ = π/4

Therefore, polar form is:

z = √2 (cos π/4 + i sin π/4)

Important Questions Class 11 Maths Chapter-Wise