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NCERT Solutions · Class 11 Maths · Chapter 4

Chapter 4: Complex Numbers and Quadratic Equations (Algebra)

Step-by-step solutions to all exercises of Chapter 4, Complex Numbers and Quadratic Equations (NCERT Class 11 Maths, 2026-27): algebra of complex numbers, powers of i, multiplicative inverse, modulus and conjugate, and the Miscellaneous Exercise. All 28 questions are answered, with the key answer highlighted.

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$i^{2} = - 1$ , $i^{4} = 1$ . $\frac{1}{a + ib} = \frac{a - ib}{a ^{2} + b ^{2}}$ . $∣ z ∣ = a^{2} + b^{2}$ , $z \overset{z}{ˉ} = ∣ z ∣^{2}$ , $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ .

Exercise 4.1

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

Solution

−3i² = 3 + i0.

3 + i0

Express in the form a + ib: i⁹ + i¹⁹

Solution

i⁹ = i, i¹⁹ = i¹⁶i³ = −i: sum = 0 + i0.

0 + i0

Express in the form a + ib: i⁻³⁹

Solution

i⁻³⁹ = i⁻⁴⁰ i = i: 0 + i1.

0 + i1

Express in the form a + ib: 3(7 + 7i) + i(7 + 7i)

Solution

21 + 21i + 7i + 7i² = 14 + 28i.

14 + 28i

Express in the form a + ib: (1 − i) − (−1 + 6i)

Solution

2 − 7i.

2 − 7i

Express in the form a + ib: (1/5 + 2i/5) − (4 + 5i/2)

Solution

(1/5 − 4) + i(2/5 − 5/2) = −19/5 − 21i/10.

−19/5 − 21i/10

Express in the form a + ib: [(1/3 + 7i/3) + (4 + i/3)] − (−4/3 + i)

Solution

(1/3 + 4 + 4/3) + i(7/3 + 1/3 − 1) = 17/3 + 5i/3.

17/3 + 5i/3

Express in the form a + ib: (1 − i)⁴

Solution

(1 − i)² = −2i, so (1 − i)⁴ = (−2i)² = −4 + i0.

−4 + i0

Express in the form a + ib: (1/3 + 3i)³

Solution

1/27 + 3(1/9)(3i) + 3(1/3)(3i)² + (3i)³ = 1/27 + i − 9 − 27i = −242/27 − 26i.

−242/27 − 26i

Express in the form a + ib: (−2 − i/3)³.

Solution

−(2 + i/3)³ = −[8 + 3(4)(i/3) + 3(2)(i/3)² + (i/3)³] = −[8 + 4i − 2/3 − i/27] = −22/3 − 107i/27.

−22/3 − 107i/27

Find the multiplicative inverse of: 4 − 3i

Solution

z⁻¹ = z̄/|z|². (4 + 3i)/25 = 4/25 + 3i/25.

4/25 + 3i/25

Find the multiplicative inverse of: √5 + 3i

Solution

z⁻¹ = z̄/|z|². (√5 − 3i)/(5 + 9) = √5/14 − 3i/14.

√5/14 − 3i/14

Find the multiplicative inverse of: −i.

Solution

z⁻¹ = z̄/|z|². (i)/1 = 0 + i (check: (−i)(i) = −i² = 1).

0 + i

Express

\frac{( 3 + i 5 ) ( 3 - i 5 )}{( 3 + 2 i ) - ( 3 - i 2 )}

in the form a + ib.

Solution

Numerator = 9 − 5i² = 14; denominator = 2√2 i. 14/(2√2 i) = 7/(√2 i) = 7i/(√2 i²) = −7i/√2 = 0 − (7√2/2)i.

0 − (7√2/2)i

Miscellaneous Exercise on Chapter 4

Evaluate [i¹⁸ + (1/i)²⁵]³.

Solution

i¹⁸ = i¹⁶i² = −1; 1/i = −i, so (1/i)²⁵ = (−i)²⁵ = −i²⁵ = −i. [−1 − i]³ = −(1 + i)³ = −(1 + 3i + 3i² + i³) = −(1 + 3i − 3 − i) = −(−2 + 2i) = 2 − 2i.

2 − 2i

For any z₁, z₂, prove Re(z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂.

Solution

Let z₁ = x₁ + iy₁, z₂ = x₂ + iy₂. z₁z₂ = (x₁x₂ − y₁y₂) + i(x₁y₂ + x₂y₁). So Re(z₁z₂) = x₁x₂ − y₁y₂ = Re z₁ Re z₂ − Im z₁ Im z₂.

Re z₁ Re z₂ − Im z₁ Im z₂

Reduce

(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i})

to the standard form.

Solution

1/(1 − 4i) = (1 + 4i)/17; 2/(1 + i) = 2(1 − i)/2 = 1 − i. Difference = (1 + 4i − 17 + 17i)/17 = (−16 + 21i)/17. (3 − 4i)/(5 + i) = (3 − 4i)(5 − i)/26 = (15 − 3i − 20i + 4i²)/26 = (11 − 23i)/26. Product = (−16 + 21i)(11 − 23i)/442 = (−176 + 368i + 231i − 483i²)/442 = (307 + 599i)/442.

(307 + 599i)/442

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

Solution

Square: (x − iy)² = (a − ib)/(c − id). Take moduli: |x − iy|² = |a − ib|/|c − id|, i.e. x² + y² = √(a² + b²)/√(c² + d²). Squaring: (x² + y²)² = (a² + b²)/(c² + d²).

(x² + y²)² = (a² + b²)/(c² + d²)

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

Solution

z₁ + z₂ + 1 = 4; z₁ − z₂ + 1 = 2 − 2i. |4/(2 − 2i)| = 4/√8 = √2.

√2

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

Solution

a² + b² = |a + ib|² = |x + i|⁴/(2x² + 1)² = (√(x² + 1))⁴/(2x² + 1)² = (x² + 1)²/(2x² + 1)².

(x² + 1)²/(2x² + 1)²

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 = −3 + 4i; z̄₁ = 2 + i. (−3 + 4i)/(2 + i) = (−3 + 4i)(2 − i)/5 = (−6 + 3i + 8i + 4)/5 = (−2 + 11i)/5: Re = −2/5. (ii) z₁z̄₁ = |z₁|² = 5 (real), so 1/5 has Im = 0.

−2/5; 0

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

Solution

(x − iy)(3 + 5i) = (3x + 5y) + i(5x − 3y) = −6 + 24i. So 3x + 5y = −6 and 5x − 3y = 24. Multiply by 3 and 5 and add: 34x = 102 ⇒ x = 3; then 5y = −15 ⇒ y = −3.

x = 3; y = −3

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

Solution

(1 + i)/(1 − i) = (1 + i)²/2 = 2i/2 = i; (1 − i)/(1 + i) = −i. Difference = 2i; modulus = 2.

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

Solution

(x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³ = (x³ − 3xy²) + i(3x²y − y³). So u = x³ − 3xy², v = 3x²y − y³; u/x + v/y = x² − 3y² + 3x² − y² = 4(x² − y²).

4(x² − y²)

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

Solution

|β| = 1 ⇒ ββ̄ = 1. Then 1 − ᾱβ = ββ̄ − ᾱβ = β(β̄ − ᾱ), so |1 − ᾱβ| = |β||conj(β − α)| = 1 × |β − α|. Hence the value is 1.

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

Solution

|1 − i| = √2: (√2)ˣ = 2ˣ ⇒ 2^(x/2) = 2ˣ ⇒ x/2 = x ⇒ x = 0. There is no non-zero integral solution: the number is 0.

no; 0

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

Solution

Take moduli: |a + ib||c + id||e + if||g + ih| = |A + iB| (modulus of a product = product of moduli). Squaring gives (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².

(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B²

If ((1 + i)/(1 − i))ᵐ = 1, find the least positive integral value of m.

Solution

(1 + i)/(1 − i) = i, so iᵐ = 1 ⇒ m is a multiple of 4: least m = 4.

m = 4

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