Exercise 1.4 Class 10 Maths Chapter 1 (Complex Numbers) — complete step-by-step solutions for the new book. Free PDF download, Al Huda Science Academy.
Exercise 1.4 Class 10 Maths — Chapter 1 (Complex Numbers) Full Solution
By Sir Usman Adal | Al Huda Science Academy | New Book (10th Mathematics)
The complete, step-by-step solution of Exercise 1.4, Chapter 1 (Complex Numbers) for Class 10 Mathematics — New Book. It covers finding the real and imaginary parts of complex expressions (including powers and inverses), and solving simultaneous linear equations with complex coefficients.
Q.1 — Find the Real and Imaginary Parts of the Following
(i) (8 − 3i)²
= 64 − 48i − 9 = 55 − 48i Real Part = 55, Imaginary Part = −48
(ii) (5 + 3i)⁻¹
= (5 − 3i)/34 Real Part = 5/34, Imaginary Part = −3/34
(iii) (4 − 5i)⁻¹
= (4 + 5i)/41 Real Part = 4/41, Imaginary Part = 5/41
(iv) (4 − 3i)⁻²
= (7 + 24i)/625 Real Part = 7/625, Imaginary Part = 24/625
(v) ((3 + 2i)/(4 + 3i))⁻¹
= (18 + i)/13 Real Part = 18/13, Imaginary Part = 1/13
(vi) ((2 − i)/(2 + i))⁻²
= (−7 + 24i)/25 Real Part = −7/25, Imaginary Part = 24/25
(vii) ((1 − 2i)/(1 + i))²
Simplify the fraction first: (1 − 2i)/(1 + i) = (−1 − 3i)/2 Then square: (−1 − 3i)²/4 = (−8 + 6i)/4 = −2 + (3/2)i Real Part = −2, Imaginary Part = 3/2
Q.2 — Solve the Simultaneous Linear Equations with Complex Coefficients for w and z
(i) 3z + (2 + i)w = 11 − i and (2 − i)z − w = −1 + i
From equation 2: w = (2 − i)z + 1 − i. Substituting into equation 1 and using (2+i)(2−i) = 5: 8z + 3 − i = 11 − i → 8z = 8 → z = 1 Then w = (2 − i)(1) + 1 − i = 3 − 2i z = 1, w = 3 − 2i
(ii) 2z + (3 + i)w = 9 − i and −iz − iw = −1 + i
From equation 2: z + w = −1 − i. Substituting into equation 1: w(1 + i) = 11 + i → w = (11 + i)/(1 + i) = 6 − 5i Then z = −1 − i − w = −7 + 4i z = −7 + 4i, w = 6 − 5i
(iii) z − 4w = 3i and 2z + 3w = 11 − 5i
From equation 1: z = 4w + 3i. Substituting into equation 2: 11w = 11 − 11i → w = 1 − i Then z = 4(1 − i) + 3i = 4 − i z = 4 − i, w = 1 − i
(iv) z + w = 3i and 2z + 3w = 2
From equation 1: z = 3i − w. Substituting into equation 2: w = 2 − 6i Then z = 3i − (2 − 6i) = −2 + 9i z = −2 + 9i, w = 2 − 6i
(v) 2z + (3 − i)w = 1 and z − (1 − i)w = 2
From equation 2: z = 2 + (1 − i)w. Substituting into equation 1: (1 + 3i)w = 5 → w = 5/(1 + 3i) = 1/2 − (3/2)i Then z = 2 − (1 − i)w = −1 + 2i z = −1 + 2i, w = 1/2 − (3/2)i
Key Facts — Exercise 1.4
Powers of i: i⁰ = 1, i¹ = i, i² = −1, i³ = −i, i⁴ = 1
Conjugate: if z = a + bi, then z̄ = a − bi
Important identities:
- (a + bi)(a − bi) = a² + b²
- (a + bi)² = (a² − b²) + 2abi
- (a − bi)² = (a² − b²) − 2abi
Inverse of a complex number: if z = a + bi (a² + b² ≠ 0), then 1/z = (a − bi)/(a² + b²)
Real & imaginary part: for z = a + bi, Real Part = a, Imaginary Part = b
General rules:
z₁/z₂ = (z₁·z̄₂)/(z₂·z̄₂)
z₁ + z₂ = (a + c) + (b + d)i
z₁ − z₂ = (a − c) + (b − d)i
z₁z₂ = (ac − bd) + (ad + bc)i
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Frequently Asked Questions
How do you find the real and imaginary part of the inverse of a complex number?
Multiply the numerator and denominator by the conjugate of the complex number, which turns the denominator into a real number (a² + b²), then read off the real and imaginary parts.
How do you solve simultaneous equations with complex coefficients?
Use the same substitution or elimination method as with real coefficients, but simplify carefully using i² = −1 whenever complex numbers are multiplied together.
What is Exercise 1.4 of Class 10 Maths Chapter 1 about?
It covers finding the real and imaginary parts of complex expressions involving powers, inverses, and quotients, and solving systems of two linear equations with complex coefficients for two unknowns.
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