NCERT Solutions for Class 11 Maths Chapter 4: Complex Numbers and Quadratic Equations – Free PDF Download

Chapter 4 of Class 11 Maths, Complex Numbers and Quadratic Equations, extends the real number system to complex numbers so that equations like x² + 1 = 0 finally have a solution. It covers the algebra of complex numbers (addition, subtraction, multiplication, division), powers of i, the modulus and conjugate of a complex number, and their representation on the Argand plane. As per the 2026-27 rationalised NCERT textbook, this chapter has a single exercise, Exercise 4.1, followed by the Miscellaneous Exercise. Below are complete, step-by-step solutions to every question. These Class 11 Maths Chapter 4 solutions are also useful as quick revision notes before exams.

Exercise 4.1

Q1. (5i)(−³⁄₅i)

(5i)(−³⁄₅i) = 5 × (−³⁄₅) × i × i = −3i².
Since i² = −1, this becomes −3(−1) = 3.
Answer: 3 (i.e., 3 + i0)

Q2. i⁹ + i¹⁹

i⁹ = i^4×2+1 = i
i¹⁹ = i^4×4+3 = i³ = −i
So i⁹ + i¹⁹ = i + (−i) = 0.
Answer: 0

Q3. i⁻³⁹

Since i⁴ = 1, we can add any multiple of 4 to the exponent without changing the value.
i⁻³⁹ = i⁻³⁹⁺⁴⁰ = i¹ = i.
Answer: i

Q4. 3(7 + i7) + i(7 + i7)

= 21 + 21i + 7i + 7i²
= 21 + 28i + 7(−1)
= 21 − 7 + 28i = 14 + 28i.
Answer: 14 + 28i

Q5. (1 − i) − (−1 + i6)

= 1 − i + 1 − 6i = 2 − 7i.
Answer: 2 − 7i

Q6. (¹⁄₅ + i²⁄₅) − (4 + i⁵⁄₂)

Real part: ¹⁄₅ − 4 = ¹⁄₅ − ²⁰⁄₅ = −¹⁹⁄₅.
Imaginary part: ²⁄₅ − ⁵⁄₂ = ⁴⁄₁₀ − ²⁵⁄₁₀ = −²¹⁄₁₀.
Answer: −¹⁹⁄₅ − i²¹⁄₁₀

Q7. [(¹⁄₃ + i⁷⁄₃) + (4 + i¹⁄₃)] − (−⁴⁄₃ + i)

First combine the bracket: (¹⁄₃ + 4) + i(⁷⁄₃ + ¹⁄₃) = ¹³⁄₃ + i⁸⁄₃.
Now subtract (−⁴⁄₃ + i):
Real: ¹³⁄₃ − (−⁴⁄₃) = ¹³⁄₃ + ⁴⁄₃ = ¹⁷⁄₃.
Imaginary: ⁸⁄₃ − 1 = ⁵⁄₃.
Answer: ¹⁷⁄₃ + i⁵⁄₃

Q8. (1 − i)⁴

First find (1 − i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i.
Then (1 − i)⁴ = [(1 − i)²]² = (−2i)² = 4i² = −4.
Answer: −4

Q9. (¹⁄₃ + 3i)³

Using (x + y)³ = x³ + 3x²y + 3xy² + y³ with x = ¹⁄₃, y = 3i:
x³ = ¹⁄₂₇
3x²y = 3 × ¹⁄₉ × 3i = i
3xy² = 3 × ¹⁄₃ × (3i)² = 1 × (9i²) = 1 × (−9) = −9
y³ = (3i)³ = 27i³ = 27(−i) = −27i
Sum: (¹⁄₂₇ − 9) + i(1 − 27) = (¹⁄₂₇ − ²⁴³⁄₂₇) − 26i = −²⁴²⁄₂₇ − 26i.
Answer: −²⁴²⁄₂₇ − 26i

Q10. (−2 − ¹⁄₃i)³

Using (x + y)³ = x³ + 3x²y + 3xy² + y³ with x = −2, y = −¹⁄₃i:
x³ = −8
3x²y = 3 × 4 × (−¹⁄₃i) = −4i
3xy² = 3 × (−2) × (−¹⁄₃i)² = −6 × (−¹⁄₉) = ²⁄₃  [since (−¹⁄₃i)² = ¹⁄₉i² = −¹⁄₉]
y³ = (−¹⁄₃i)³ = −¹⁄₂₇i³ = −¹⁄₂₇(−i) = ¹⁄₂₇i
Sum: real part = −8 + ²⁄₃ = −²⁴⁄₃ + ²⁄₃ = −²²⁄₃.
Imaginary part = −4 + ¹⁄₂₇ = −¹⁰⁸⁄₂₇ + ¹⁄₂₇ = −¹⁰⁷⁄₂₇.
Answer: −²²⁄₃ − ¹⁰⁷⁄₂₇i

Find the multiplicative inverse of each of the complex numbers given in Q11 to Q13.

Q11. 4 − 3i

Let z = 4 − 3i. Then z = 4 + 3i and |z|² = 4² + (−3)² = 16 + 9 = 25.
Multiplicative inverse: z⁻¹ = z ⁄ |z|² = (4 + 3i) ⁄ 25 = ⁴⁄₂₅ + ³⁄₂₅i.
Answer: ⁴⁄₂₅ + ³⁄₂₅i

Q12. √5 + 3i

Let z = √5 + 3i. Then z = √5 − 3i and |z|² = (√5)² + 3² = 5 + 9 = 14.
z⁻¹ = (√5 − 3i) ⁄ 14 = ^√5⁄₁₄ − ³⁄₁₄i.
Answer: ^√5⁄₁₄ − ³⁄₁₄i

Q13. −i

Let z = −i. Then z = i and |z|² = 0² + (−1)² = 1.
z⁻¹ = i ⁄ 1 = i.
Check: (−i)(i) = −i² = 1 ✓
Answer: i

Q14. Express the following expression in the form a + ib: [(3 + i√5)(3 − i√5)] ⁄ [(√3 + √2i) − (√3 − i√2)]

Numerator: (3 + i√5)(3 − i√5) = 3² − (i√5)² = 9 − (−5) = 9 + 5 = 14.
Denominator: (√3 + √2i) − (√3 − i√2) = √3 − √3 + √2i + √2i = 2√2i.
So the expression = 14 ⁄ (2√2i) = 7 ⁄ (√2i).
Multiply numerator and denominator by −i (since 1⁄i = −i):
= 7 × (−i) ⁄ √2 = −⁷⁄_√2i = −^7√2⁄₂i  (rationalising ⁷⁄_√2 = ^7√2⁄₂).
Answer: 0 − ^7√2⁄₂i

Miscellaneous Exercise

Q1. Evaluate: [i¹⁸ + (¹⁄_i)²⁵]³

i¹⁸ = i^4×4+2 = i² = −1.
Since ¹⁄_i = −i, we get (¹⁄_i)²⁵ = (−i)²⁵ = −i²⁵ = −i^4×6+1 = −i.
So the bracket = −1 + (−i) = −1 − i = −(1 + i).
Now (1 + i)² = 1 + 2i + i² = 2i, so (1 + i)³ = (1 + i)(2i) = 2i + 2i² = −2 + 2i.
Therefore [−(1 + i)]³ = −(1 + i)³ = −(−2 + 2i) = 2 − 2i.
Answer: 2 − 2i

Q2. For any two complex numbers z₁ and z₂, prove that Re(z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂.

Let z₁ = a + i b and z₂ = c + i d, so Re z₁ = a, Im z₁ = b, Re z₂ = c, Im z₂ = d.
z₁z₂ = (a + ib)(c + id) = ac + iad + ibc + i²bd = (ac − bd) + i(ad + bc).
So Re(z₁z₂) = ac − bd = Re z₁ Re z₂ − Im z₁ Im z₂. Hence proved.

Q3. Reduce (¹⁄_1−4i − ²⁄_1+i)(³⁻⁴ⁱ⁄_5+i) to the standard form.

First, ¹⁄_1−4i = ¹⁺⁴ⁱ⁄_{(1−4i)(1+4i)} = ¹⁺⁴ⁱ⁄₁₇.
Also, ²⁄_1+i = ²⁽¹⁻ⁱ⁾⁄_(1+i)(1−i) = ²⁽¹⁻ⁱ⁾⁄₂ = 1 − i.
So ¹⁄_1−4i − ²⁄_1+i = ¹⁺⁴ⁱ⁄₁₇ − (1 − i) = ^{1+4i − 17(1−i)}⁄₁₇ = ^−16+21i⁄₁₇.
Next, simplify ³⁻⁴ⁱ⁄_5+i = ^{(3−4i)(5−i)}⁄_(5+i)(5−i) = ^{15−3i−20i+4i²}⁄₂₆ = ^15−23i−4⁄₂₆ = ¹¹⁻²³ⁱ⁄₂₆.
Now multiply: ^−16+21i⁄₁₇ × ¹¹⁻²³ⁱ⁄₂₆.
Numerator: (−16+21i)(11−23i) = −176 + 368i + 231i − 483i² = −176 + 599i + 483 = 307 + 599i.
Denominator: 17 × 26 = 442.
Answer: ³⁰⁷⁄₄₄₂ + ⁵⁹⁹⁄₄₄₂i

Q4. If x − iy = √[(a−ib)⁄(c−id)], prove that (x²+y²)² = (a²+b²)⁄(c²+d²).

Squaring the given relation: (x − iy)² = ^a−ib⁄_c−id  …(1)
Taking the complex conjugate of both sides of (1): the conjugate of (x − iy)² is (x + iy)², and the conjugate of ^a−ib⁄_c−id is ^a+ib⁄_c+id. So:
(x + iy)² = ^a+ib⁄_c+id  …(2)
Multiplying (1) and (2):
(x − iy)²(x + iy)² = ^{(a−ib)(a+ib)}⁄_{(c−id)(c+id)}
[(x − iy)(x + iy)]² = ^a²+b²⁄_c²+d²
Since (x − iy)(x + iy) = x² + y², we get (x²+y²)² = ^a²+b²⁄_c²+d². Hence proved.

Q5. If z₁ = 2 − i, z₂ = 1 + i, find |^z₁+z₂+1⁄_z1−z2+1|.

z₁ + z₂ + 1 = (2 − i) + (1 + i) + 1 = 4.
z₁ − z₂ + 1 = (2 − i) − (1 + i) + 1 = 2 − 2i.
So the ratio = ⁴⁄_2−2i = ⁴⁄_2(1−i) = ²⁄_1−i = ²⁽¹⁺ⁱ⁾⁄_(1−i)(1+i) = ²⁽¹⁺ⁱ⁾⁄₂ = 1 + i.
|1 + i| = √(1²+1²) = √2.
Answer: √2

Q6. If a + ib = ^(x+i)²⁄_2x²+1, prove that a² + b² = ^(x²+1)²⁄_(2x²+1)².

Taking modulus of both sides: |a + ib| = ^|(x+i)²|⁄_|2x²+1| = ^|x+i|²⁄_2x²+1  (since 2x²+1 is real and positive, and |z²| = |z|²).
Squaring both sides: |a + ib|² = ^|x+i|&sup4;⁄_(2x²+1)².
Now a² + b² = |a + ib|², and |x + i|² = x² + 1, so |x + i|&sup4; = (x²+1)².
Therefore a² + b² = ^(x²+1)²⁄_(2x²+1)². Hence proved.

Q7. Let z₁ = 2 − i, z₂ = −2 + i. Find:

1. Re(^z₁z₂⁄_z1)
2. Im(¹⁄_z1z1)

(i) z₁z₂ = (2−i)(−2+i) = −4 + 2i + 2i − i² = −4 + 4i + 1 = −3 + 4i.
z₁ = 2 + i.
^z₁z₂⁄_z1 = ⁻³⁺⁴ⁱ⁄_2+i = ^{(−3+4i)(2−i)}⁄_(2+i)(2−i) = ^{−6+3i+8i−4i²}⁄₅ = ^−6+11i+4⁄₅ = ⁻²⁺¹¹ⁱ⁄₅.
Re(^z₁z₂⁄_z1) = −²⁄₅.
Answer (i): −²⁄₅

(ii) z₁z₁ = (2−i)(2+i) = 4 − i² = 4 + 1 = 5 (a real number, equal to |z₁|²).
¹⁄_z1z1 = ¹⁄₅, which is purely real.
Im(¹⁄₅) = 0.
Answer (ii): 0

Q8. Find the real numbers x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.

The conjugate of −6 − 24i is −6 + 24i.
Expand (x − iy)(3 + 5i) = 3x + 5xi − 3yi − 5yi² = (3x + 5y) + i(5x − 3y).
Equating with −6 + 24i:
3x + 5y = −6  …(1)
5x − 3y = 24  …(2)
Multiply (1) by 3 and (2) by 5: 9x + 15y = −18 and 25x − 15y = 120. Adding: 34x = 102, so x = 3.
Substitute in (1): 9 + 5y = −6, so 5y = −15, y = −3.
Check in (2): 5(3) − 3(−3) = 15 + 9 = 24 ✓
Answer: x = 3, y = −3

Q9. Find the modulus of ¹⁺ⁱ⁄_1−i − ¹⁻ⁱ⁄_1+i.

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⁄_(1+i)(1−i) = ^1−2i+i²⁄₂ = ⁻²ⁱ⁄₂ = −i.
Difference: i − (−i) = 2i.
|2i| = 2.
Answer: 2

Q10. If (x + iy)³ = u + iv, then show that ^u⁄_x + ^v⁄_y = 4(x² − y²).

Expand: (x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³
= x³ + 3ix²y − 3xy² − iy³
= (x³ − 3xy²) + i(3x²y − y³).
So u = x³ − 3xy² = x(x² − 3y²) and v = 3x²y − y³ = y(3x² − y²).
Then ^u⁄_x = x² − 3y² and ^v⁄_y = 3x² − y² (for x ≠ 0, y ≠ 0).
Adding: ^u⁄_x + ^v⁄_y = (x² − 3y²) + (3x² − y²) = 4x² − 4y² = 4(x² − y²). Hence proved.

Q11. If α and β are different complex numbers with |β| = 1, then find |^β−α⁄_1−αβ|.

We compute |β−α|² and |1−αβ|² and compare them.
|β−α|² = (β−α)(β−α) = |β|² − βα − αβ + |α|² = 1 − αβ − αβ + |α|²  (using |β|²=1, and βα = αβ).
|1−αβ|² = (1−αβ)(1−αβ) = 1 − αβ − αβ + ααββ = 1 − αβ − αβ + |α|²|β|² = 1 − αβ − αβ + |α|²  (again using |β|²=1).
Both expressions are identical, so |β−α|² = |1−αβ|², which gives |β−α| = |1−αβ|.
Answer: |^β−α⁄_1−αβ| = 1

Q12. Find the number of non-zero integral solutions of the equation |1 − i|^x = 2^x.

|1 − i| = √(1²+(−1)²) = √2.
So the equation becomes (√2)^x = 2^x, i.e., 2^x⁄2 = 2^x.
Equating exponents (same base): ^x⁄₂ = x, which gives x = 0.
Since the question asks for non-zero integral solutions and x = 0 is the only solution (and it is zero), there is no non-zero integral value of x satisfying the equation.
Answer: 0 (no non-zero integral solutions exist)

Q13. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that (a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B².

Taking the modulus of both sides of the given equation and using the property |z₁z₂z₃z₄| = |z₁||z₂||z₃||z₄|:
|a+ib| · |c+id| · |e+if| · |g+ih| = |A+iB|.
Squaring both sides:
|a+ib|² · |c+id|² · |e+if|² · |g+ih|² = |A+iB|²
(a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B². Hence proved.

Q14. If (¹⁺ⁱ⁄_1−i)^m = 1, then find the least positive integral value of m.

First simplify ¹⁺ⁱ⁄_1−i = ^(1+i)²⁄_(1−i)(1+i) = ^1+2i+i²⁄₂ = ²ⁱ⁄₂ = i.
So the equation becomes i^m = 1.
Since i⁴ = 1 and 4 is the smallest positive integer for which this holds, the least positive integral value of m is 4.
Answer: m = 4

Class 11 Maths Chapter 4 – Notes and Extra Questions

Before attempting the exercises, it helps to be clear on the key definitions and formulas of this chapter:

• Imaginary unit: i = √−1, so i² = −1. A complex number is any number of the form a + ib, where a and b are real numbers; a is called the real part and b the imaginary part.
• Algebra of complex numbers: for z₁ = a + ib and z₂ = c + id:
  • Addition: z₁ + z₂ = (a+c) + i(b+d)
  • Multiplication: z₁z₂ = (ac−bd) + i(ad+bc)
  • Division (z₂ ≠ 0): ^z₁⁄_z2 is found by multiplying numerator and denominator by the conjugate of z₂
• Powers of i: i¹=i, i²=−1, i³=−i, i⁴=1, and the cycle repeats. In general, for any integer k: i^4k=1, i^4k+1=i, i^4k+2=−1, i^4k+3=−i. Also ¹⁄_i = −i.
• Modulus: for z = a + ib, |z| = √(a²+b²), the distance of the point (a, b) from the origin on the Argand plane.
• Conjugate: the conjugate of z = a + ib is z = a − ib. Useful identities: z + z = 2a (real), z·z = a²+b² = |z|² (real and non-negative).
• Multiplicative inverse: for a non-zero z = a + ib, z⁻¹ = ^z⁄_|z|² = ^a⁄_a²+b² − ^b⁄_a²+b²i.
• Argand plane: a complex number a + ib is plotted as the point (a, b), with the horizontal axis as the real axis and the vertical axis as the imaginary axis. The modulus |z| is the distance of this point from the origin.
• Handy algebraic shortcuts: (a+ib)(a−ib) = a²+b²; |z₁z₂| = |z₁||z₂|; Re(z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂. These are frequently used to simplify or prove results quickly.

Frequently Asked Questions

How many exercises are there in Class 11 Maths Chapter 4?
As per the 2026-27 rationalised NCERT textbook, Chapter 4 (Complex Numbers and Quadratic Equations) has one exercise, Exercise 4.1 (14 questions), followed by the Miscellaneous Exercise (14 questions) — a total of 28 questions.

Does the current NCERT syllabus for this chapter include solving quadratic equations with complex roots?
In the rationalised (post-2023) edition, the detailed treatment of quadratic equations with complex roots and the polar/modulus-argument form has been removed from the main exercises; the chapter now focuses on the algebra of complex numbers, powers of i, modulus, conjugate, and the Argand plane, tested through Exercise 4.1 and the Miscellaneous Exercise.

What is the most important formula in this chapter?
The relation i² = −1 is the foundation of the entire chapter. Almost every simplification in Exercise 4.1 and the Miscellaneous Exercise relies on correctly substituting i² = −1 while expanding products or powers.

What is the easiest way to divide two complex numbers?
Multiply both the numerator and the denominator by the conjugate of the denominator. This converts the denominator into a real number (a²+b²), after which the expression can be separated into real and imaginary parts easily.