NCERT Solutions for Class 11 Maths Chapter 4: Complex Numbers and Quadratic Equations (2026–27)

These Class 11 Maths Chapter 4 solutions cover Complex Numbers and Quadratic Equations from the NCERT textbook (Reprint 2026–27). Every question of Exercise 4.1 and the Miscellaneous Exercise on Chapter 4 is reproduced verbatim and solved step by step — expressing numbers in the form a + ib, finding multiplicative inverses, moduli and conjugates — with each answer cross-checked against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 4 Exercises: 4.1, Miscellaneous Session: 2026–27
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Chapter 4 Overview

Chapter 4, Complex Numbers and Quadratic Equations, extends the real number system so that equations such as x2 + 1 = 0 have solutions. It introduces the imaginary unit i = √(−1) and the complex number a + ib, then develops the algebra of complex numbers — addition, subtraction, multiplication and division — together with the powers of i, the square roots of negative numbers, the modulus and conjugate, the multiplicative inverse, and the geometric picture in the Argand plane. The solutions below work through every part of Exercise 4.1 and the Miscellaneous Exercise, all reduced to the standard form a + ib.

Key Concepts & Definitions

Imaginary unit: i = √(−1), so i2 = −1. It is a solution of x2 + 1 = 0.

Complex number: a number of the form z = a + ib where a, b are real. Here a = Re z (real part) and b = Im z (imaginary part).

Equality: a + ib = c + id if and only if a = c and b = d.

Conjugate: the conjugate of z = a + ib is z̄ = a − ib (reflection in the real axis).

Modulus: |z| = √(a2 + b2), the distance of the point (a, b) from the origin in the Argand plane.

Multiplicative inverse: for z ≠ 0, z−1 = z̄ / |z|2 = (a − ib)/(a2 + b2).

Square root of a negative number: for a > 0, √(−a) = √a·i. Note √a·√b = √(ab) fails when both a and b are negative.

Important Formulas (Chapter 4)

Powers of i: for any integer k, i4k = 1, i4k+1 = i, i4k+2 = −1, i4k+3 = −i. Also i−1 = −i.

Sum / product: (a + ib) + (c + id) = (a + c) + i(b + d); (a + ib)(c + id) = (ac − bd) + i(ad + bc).

Division (rationalise by the conjugate): (a + ib)/(c + id) = [(a + ib)(c − id)] / (c2 + d2).

Conjugate & modulus: z̄ = a − ib, |z| = √(a2 + b2), z·z̄ = |z|2.

Multiplicative inverse: z−1 = (a − ib)/(a2 + b2).

Properties: |z1z2| = |z1||z2|, |z1/z2| = |z1|/|z2|, and (z1z2)̄ = z̄1z̄2.

Exercise 4.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

1. (5i)(−3/5 i)

SOLUTION (5i)(−3/5 i) = 5 × (−3/5) × i2 = −3 × i2. Since i2 = −1, this is −3 × (−1) = 3. ∴ the result is 3 + i0.

2. i9 + i19

SOLUTION i9 = i4×2 + 1 = i, and i19 = i4×4 + 3 = i3 = −i. Sum = i + (−i) = 0. ∴ the result is 0 + i0.

3. i−39

SOLUTION i−39 = 1/i39. Now i39 = i4×9 + 3 = i3 = −i. So i−39 = 1/(−i) = −1/i = −1/i × i/i = −i/i2 = −i/(−1) = i. ∴ the result is 0 + i1.

4. 3(7 + i7) + i (7 + i7)

SOLUTION 3(7 + i7) = 21 + 21i. i(7 + i7) = 7i + 7i2 = 7i − 7 = −7 + 7i. Adding: (21 − 7) + (21 + 7)i = 14 + 28i. ∴ the result is 14 + 28i.

5. (1 – i) – ( –1 + i6)

SOLUTION (1 − i) − (−1 + 6i) = 1 − i + 1 − 6i. = (1 + 1) + (−1 − 6)i = 2 − 7i. ∴ the result is 2 − 7i.

6. (1/5 + i 2/5) − (4 + i 5/2)

SOLUTION Real part: 1/5 − 4 = (1 − 20)/5 = −19/5. Imaginary part: 2/5 − 5/2 = (4 − 25)/10 = −21/10. ∴ the result is −19/5 − (21/10)i.

7. [(1/3 + i 7/3) + (4 + i 1/3)] − (−4/3 + i)

SOLUTION First add the two brackets: real 1/3 + 4 = 13/3; imaginary 7/3 + 1/3 = 8/3. This gives 13/3 + i(8/3). Now subtract (−4/3 + i): real 13/3 − (−4/3) = 13/3 + 4/3 = 17/3. Imaginary 8/3 − 1 = (8 − 3)/3 = 5/3. ∴ the result is 17/3 + i(5/3).

8. (1 – i)4

SOLUTION First (1 − i)2 = 1 − 2i + i2 = 1 − 2i − 1 = −2i. Then (1 − i)4 = [(1 − i)2]2 = (−2i)2 = 4i2 = −4. ∴ the result is −4 + i0.

9. (1/3 + 3i)3

SOLUTION Use (a + b)3 = a3 + 3a2b + 3ab2 + b3 with a = 1/3, b = 3i. a3 = 1/27; 3a2b = 3 × (1/9) × 3i = i; 3ab2 = 3 × (1/3) × 9i2 = 9i2 = −9; b3 = 27i3 = −27i. Sum: (1/27 − 9) + (1 − 27)i = (1 − 243)/27 + (−26)i = −242/27 − 26i. ∴ the result is −242/27 − 26i.

10. (−2 − 1/3 i)3

SOLUTION Write it as −(2 + i/3)3. Use (a + b)3 with a = 2, b = i/3. a3 = 8; 3a2b = 3 × 4 × i/3 = 4i; 3ab2 = 3 × 2 × (i2/9) = 6 × (−1/9) = −2/3; b3 = i3/27 = −i/27. (2 + i/3)3 = (8 − 2/3) + (4 − 1/27)i = 22/3 + (107/27)i. Therefore (−2 − i/3)3 = −[22/3 + (107/27)i] = −22/3 − (107/27)i. ∴ the result is −22/3 − (107/27)i.

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i

SOLUTION For z = 4 − 3i, z̄ = 4 + 3i and |z|2 = 42 + (−3)2 = 16 + 9 = 25. z−1 = z̄/|z|2 = (4 + 3i)/25 = 4/25 + (3/25)i. ∴ the inverse is 4/25 + (3/25)i.

12. 5 + 3i

SOLUTION For z = 5 + 3i, z̄ = 5 − 3i and |z|2 = 52 + 32 = 25 + 9 = 34. z−1 = (5 − 3i)/34 = 5/34 − (3/34)i. ∴ the inverse is 5/34 − (3/34)i.

13. – i

SOLUTION For z = −i = 0 − i, z̄ = i and |z|2 = 02 + (−1)2 = 1. z−1 = z̄/|z|2 = i/1 = i. ∴ the inverse is 0 + i1 (that is, i).

14. Express the following expression in the form of a + ib : [(3 + i√5)(3 − i√5)] / [(√3 + i√2) − (√3 − i√2)]

SOLUTION Numerator: (3 + i√5)(3 − i√5) = 32 − (i√5)2 = 9 − i2(5) = 9 + 5 = 14. Denominator: (√3 + i√2) − (√3 − i√2) = 2i√2. Expression = 14 / (2√2 i) = 7/(√2 i). Multiply top and bottom by i: = 7i/(√2 i2) = 7i/(−√2) = −(7/√2)i. Rationalising, 7/√2 = 7√2/2, so the value is −(7√2/2)i. ∴ the result is 0 − (7√2/2)i.

Miscellaneous Exercise on Chapter 4 — Solutions

1. Evaluate: [i18 + (1/i)25]3.

SOLUTION i18 = i4×4 + 2 = i2 = −1. (1/i)25 = i−25 = 1/i25; i25 = i4×6 + 1 = i, so i−25 = 1/i = −i. Inside the bracket: −1 + (−i) = −(1 + i). Cube it: [−(1 + i)]3 = −(1 + i)3. Now (1 + i)2 = 1 + 2i + i2 = 2i, so (1 + i)3 = (1 + i)(2i) = 2i + 2i2 = −2 + 2i. Therefore the value = −(−2 + 2i) = 2 − 2i. ∴ the result is 2 − 2i.

2. For any two complex numbers z1 and z2, prove that Re (z1z2) = Re z1 Re z2 − Imz1 Imz2.

SOLUTION Let z1 = a + ib and z2 = c + id, so Re z1 = a, Im z1 = b, Re z2 = c, Im z2 = d. z1z2 = (a + ib)(c + id) = (ac − bd) + i(ad + bc). Hence Re(z1z2) = ac − bd = (Re z1)(Re z2) − (Im z1)(Im z2). ∴ Re(z1z2) = Re z1 Re z2 − Im z1 Im z2. Proved.

3. Reduce [1/(1 − 4i) − 2/(1 + i)] [(3 − 4i)/(5 + i)] to the standard form.

SOLUTION First bracket: combine over (1 − 4i)(1 + i) = 1 + i − 4i − 4i2 = 5 − 3i. Numerator: 1(1 + i) − 2(1 − 4i) = 1 + i − 2 + 8i = −1 + 9i. So the bracket = (−1 + 9i)/(5 − 3i). Multiply by (3 − 4i)/(5 + i): product = [(−1 + 9i)(3 − 4i)] / [(5 − 3i)(5 + i)]. Top: (−1 + 9i)(3 − 4i) = −3 + 4i + 27i − 36i2 = −3 + 31i + 36 = 33 + 31i. Bottom: (5 − 3i)(5 + i) = 25 + 5i − 15i − 3i2 = 25 − 10i + 3 = 28 − 10i. So value = (33 + 31i)/(28 − 10i). Multiply by the conjugate (28 + 10i): denominator = 282 + 102 = 784 + 100 = 884. Numerator: (33 + 31i)(28 + 10i) = 924 + 330i + 868i + 310i2 = 924 + 1198i − 310 = 614 + 1198i. Value = (614 + 1198i)/884 = (307 + 599i)/442. ∴ standard form is 307/442 + (599/442)i.

4. If x − iy = √[(a − ib)/(c − id)] prove that (x2 + y2)2 = (a2 + b2)/(c2 + d2).

SOLUTION Given x − iy = √[(a − ib)/(c − id)]. Squaring: (x − iy)2 = (a − ib)/(c − id). Take the conjugate of both sides: (x + iy)2 = (a + ib)/(c + id). Multiply the two equations: (x − iy)2(x + iy)2 = [(a − ib)(a + ib)] / [(c − id)(c + id)]. Left side = [(x − iy)(x + iy)]2 = (x2 + y2)2. Right side = (a2 + b2)/(c2 + d2). ∴ (x2 + y2)2 = (a2 + b2)/(c2 + d2). Proved.

5. If z1 = 2 – i, z2 = 1 + i, find |(z1 + z2 + 1)/(z1 − z2 + 1)|.

SOLUTION z1 + z2 + 1 = (2 − i) + (1 + i) + 1 = 4 + 0i = 4. z1 − z2 + 1 = (2 − i) − (1 + i) + 1 = 2 − 2i. So the expression = 4/(2 − 2i) = 4/[2(1 − i)] = 2/(1 − i) = 2(1 + i)/[(1 − i)(1 + i)] = 2(1 + i)/2 = 1 + i. Its modulus: |1 + i| = √(12 + 12) = √2. ∴ the modulus is √2.

6. If a + ib = (x + i)2/(2x2 + 1), prove that a2 + b2 = (x2 + 1)2/(2x2 + 1)2.

SOLUTION a + ib = (x + i)2/(2x2 + 1) = (x2 + 2xi + i2)/(2x2 + 1) = (x2 − 1 + 2xi)/(2x2 + 1). So a = (x2 − 1)/(2x2 + 1) and b = 2x/(2x2 + 1). a2 + b2 = [(x2 − 1)2 + (2x)2] / (2x2 + 1)2 = [x4 − 2x2 + 1 + 4x2] / (2x2 + 1)2. = (x4 + 2x2 + 1)/(2x2 + 1)2 = (x2 + 1)2/(2x2 + 1)2. ∴ a2 + b2 = (x2 + 1)2/(2x2 + 1)2. Proved.

7. Let z1 = 2 – i, z2 = −2 + i. Find (i) Re [(z1z2)/z̄1],   (ii) Im [1/(z1z̄1)].

SOLUTION (i) z1z2 = (2 − i)(−2 + i) = −4 + 2i + 2i − i2 = −4 + 4i + 1 = −3 + 4i. z̄1 = 2 + i. So (z1z2)/z̄1 = (−3 + 4i)/(2 + i). Multiply by (2 − i): denominator = 22 + 12 = 5. Numerator: (−3 + 4i)(2 − i) = −6 + 3i + 8i − 4i2 = −6 + 11i + 4 = −2 + 11i. So the quotient = (−2 + 11i)/5 = −2/5 + (11/5)i. Hence Re = −2/5. (ii) z1z̄1 = |z1|2 = 22 + (−1)2 = 5, a real number. So 1/(z1z̄1) = 1/5 = 1/5 + 0i. Therefore Im [1/(z1z̄1)] = 0.

8. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

SOLUTION The conjugate of −6 − 24i is −6 + 24i. Expand: (x − iy)(3 + 5i) = 3x + 5xi − 3yi − 5yi2 = (3x + 5y) + (5x − 3y)i. Equate to −6 + 24i: 3x + 5y = −6 and 5x − 3y = 24. Solve: from these, multiply (1) by 3 and (2) by 5: 9x + 15y = −18, 25x − 15y = 120; adding gives 34x = 102, so x = 3. Then 3(3) + 5y = −6 ⇒ 5y = −15 ⇒ y = −3. ∴ x = 3, y = −3.

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

SOLUTION (1 + i)/(1 − i) = (1 + i)2/[(1 − i)(1 + i)] = (1 + 2i + i2)/2 = (2i)/2 = i. (1 − i)/(1 + i) = (1 − i)2/2 = (1 − 2i + i2)/2 = (−2i)/2 = −i. Difference: i − (−i) = 2i. Modulus = |2i| = 2. ∴ the modulus is 2.

10. If (x + iy)3 = u + iv, then show that u/x + v/y = 4(x2 − y2).

SOLUTION (x + iy)3 = x3 + 3x2(iy) + 3x(iy)2 + (iy)3 = x3 + 3x2yi − 3xy2 − y3i. = (x3 − 3xy2) + i(3x2y − y3). So u = x3 − 3xy2 and v = 3x2y − y3. u/x = (x3 − 3xy2)/x = x2 − 3y2;   v/y = (3x2y − y3)/y = 3x2 − y2. u/x + v/y = (x2 − 3y2) + (3x2 − y2) = 4x2 − 4y2 = 4(x2 − y2). ∴ u/x + v/y = 4(x2 − y2). Proved.

11. If α and β are different complex numbers with |β| = 1, then find |(β − α)/(1 − ᾱβ)|.

SOLUTION Since |β| = 1, we have β̄β = |β|2 = 1, so 1 = β̄β. Write the denominator using this: 1 − ᾱβ = β̄β − ᾱβ = β(β̄ − ᾱ) = β·(β − α)̄. So |(β − α)/(1 − ᾱβ)| = |β − α| / [|β|·|(β − α)̄|] = |β − α| / (1 × |β − α|) = 1. (Using |z̄| = |z| and |β| = 1, with β ≠ α so the modulus is non-zero.) ∴ the value is 1.

12. Find the number of non-zero integral solutions of the equation |1 − i|x = 2x.

SOLUTION |1 − i| = √(12 + (−1)2) = √2 = 21/2. So the equation is (21/2)x = 2x, i.e. 2x/2 = 2x. Equating exponents: x/2 = x ⇒ x = 2x ⇒ x = 0. The only solution is x = 0, which is not a non-zero integer. ∴ the number of non-zero integral solutions is 0.

13. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

SOLUTION Take the modulus of both sides. The modulus of a product is the product of moduli, so |a + ib|·|c + id|·|e + if|·|g + ih| = |A + iB|. That is √(a2 + b2) √(c2 + d2) √(e2 + f2) √(g2 + h2) = √(A2 + B2). Squaring both sides removes all the square roots: (a2 + b2)(c2 + d2)(e2 + f2)(g2 + h2) = A2 + B2. Proved.

14. If [(1 + i)/(1 – i)]m = 1, then find the least positive integral value of m.

SOLUTION First simplify the base: (1 + i)/(1 − i) = (1 + i)2/[(1 − i)(1 + i)] = (2i)/2 = i. So the equation becomes im = 1. Powers of i cycle: i1 = i, i2 = −1, i3 = −i, i4 = 1. The smallest positive m giving 1 is m = 4. ∴ the least positive integral value is m = 4.

Common Mistakes to Avoid

Watch out for these

  • Forgetting that i2 = −1; this sign flip is the source of most errors when expanding products and powers.
  • Applying √a·√b = √(ab) when both a and b are negative — it is invalid, e.g. √(−1)·√(−1) ≠ √1.
  • For division, not multiplying by the conjugate of the denominator to make it real.
  • Confusing the multiplicative inverse z−1 = z̄/|z|2 with the conjugate z̄ itself.
  • Reducing powers of i incorrectly — reduce the exponent modulo 4 (i4k = 1).
  • Leaving an answer with i in the denominator; always rationalise to the form a + ib.

Practice MCQs & Assertion–Reason

1. The value of i9 + i19 is:

(a) 0    (b) i    (c) −i    (d) 2i

2. The multiplicative inverse of −i is:

(a) i    (b) −i    (c) 1    (d) −1

3. The modulus of the complex number 3 + 4i is:

(a) 5    (b) 7    (c) 25    (d) √7

4. (1 − i)4 equals:

(a) 4    (b) −4    (c) 4i    (d) −4i

5. The conjugate of the complex number 2 − 5i is:

(a) 2 + 5i    (b) −2 + 5i    (c) −2 − 5i    (d) 5 − 2i

6. The value of (1 + i)/(1 − i) is:

(a) 1    (b) −1    (c) i    (d) −i

7. If z = a + ib, then z·z̄ equals:

(a) a2 − b2    (b) a2 + b2    (c) 2ab    (d) a + b

8. The least positive integer m for which [(1 + i)/(1 − i)]m = 1 is:

(a) 1    (b) 2    (c) 3    (d) 4

9. The real part of the complex number −3 + 4i is:

(a) 4    (b) −3    (c) 3    (d) −4

10. The number of non-zero integral solutions of |1 − i|x = 2x is:

(a) 0    (b) 1    (c) 2    (d) infinite

Answer key: 1-(a), 2-(b), 3-(a), 4-(b), 5-(a), 6-(c), 7-(b), 8-(d), 9-(b), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The multiplicative inverse of 4 − 3i is 4/25 + (3/25)i.

Reason: For z ≠ 0, z−1 = z̄/|z|2.

A-R 2. Assertion: |1 − i| = √2.

Reason: The modulus of a + ib is √(a2 + b2).

A-R 3. Assertion: √(−2) × √(−3) = √6.

Reason: √a × √b = √(ab) holds for all real numbers a and b.

A-R 4. Assertion: For z = a + ib, z·z̄ is always a non-negative real number.

Reason: z·z̄ = a2 + b2 = |z|2.

A-R 5. Assertion: i4k = 1 for every integer k.

Reason: The powers of i repeat in a cycle of length 4.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • i = √(−1), i2 = −1; powers of i repeat every 4: i4k = 1, i4k+1 = i, i4k+2 = −1, i4k+3 = −i.
  • A complex number is z = a + ib with Re z = a, Im z = b; equality means equal real and equal imaginary parts.
  • Conjugate z̄ = a − ib; modulus |z| = √(a2 + b2); and z·z̄ = |z|2.
  • Multiplicative inverse: z−1 = z̄/|z|2 = (a − ib)/(a2 + b2).
  • Divide by multiplying numerator and denominator by the conjugate of the denominator.
  • |z1z2| = |z1||z2| and |z̄| = |z| — very useful in modulus-of-product proofs.
  • √a·√b = √(ab) is invalid when both a and b are negative.

How to score full marks in this chapter

Reduce every answer to the standard form a + ib and never leave i in the denominator — rationalise using the conjugate. For modulus questions, look for chances to use |z1z2| = |z1||z2| and |z̄| = |z| instead of expanding fully. When reducing powers of i, divide the exponent by 4 and use only the remainder. Write each algebraic step on its own line so partial marks are easy to earn, and check the sign carefully every time i2 appears.

Frequently Asked Questions

What is Class 11 Maths Chapter 4 about?

Chapter 4, Complex Numbers and Quadratic Equations, introduces the imaginary unit i = √(−1) and the complex number a + ib. It covers the algebra of complex numbers (addition, multiplication, division), powers of i, square roots of negative numbers, the modulus and conjugate, the multiplicative inverse, and the Argand plane.

How many exercises are there in Class 11 Maths Chapter 4?

There is one numbered exercise, Exercise 4.1 (14 questions), followed by a Miscellaneous Exercise on Chapter 4 (14 questions). Both are solved in full on this page.

How do you find the multiplicative inverse of a complex number?

For a non-zero complex number z = a + ib, the multiplicative inverse is z−1 = z̄/|z|2 = (a − ib)/(a2 + b2). For example, the inverse of 4 − 3i is (4 + 3i)/25 = 4/25 + (3/25)i.

Are these Class 11 Maths Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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