Exercise 1.2 Class 10 Maths Chapter 1 (Complex Numbers) — complete step-by-step solutions for the new book. Free PDF download, Al Huda Science Academy.
Exercise 1.2 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 1.2, Chapter 1 (Complex Numbers) for Class 10 Mathematics — New Book. It covers addition, subtraction, multiplication, and division of complex numbers, additive and multiplicative inverses, verification of algebraic properties, and solving for unknowns.
Q.1 — Simplify and Write in the Form a + bi
(i) (2 + 5i) + (3 − 2i)
= 5 + 3i
(ii) (16 − 3i) + (9 + 2i)
= 25 − i
(iii) (9 − 2i) − (7 − 3i)
= 2 + i
(iv) (11 + 9i) − (9 − 7i)
= 2 + 16i
(v) (3 + 4i)(2 − 3i)
= 6 − 9i + 8i − 12i² = 6 − i + 12 = 18 − i
(vi) (5 − 2i)(3 − 4i)
= 15 − 20i − 6i + 8i² = 15 − 26i − 8 = 7 − 26i
(vii) (3 − 5i) ÷ (2 − 4i)
Multiply by conjugate (2 + 4i): = (26 + 2i)/20 = 13/10 + (1/10)i
(viii) (5 + 2i) ÷ (6 − 3i)
Multiply by conjugate (6 + 3i): = (24 + 27i)/45 = 8/15 + (3/5)i
Q.2 — Additive Inverse of Each Complex Number
(i) 3 + 2i → −3 − 2i
(ii) 4 − 3i → −4 + 3i
(iii) 5 − 7i → −5 + 7i
(iv) −2/3 + (5/4)i → 2/3 − (5/4)i
Q.3 — Multiplicative Inverse of Each Complex Number
(i) 4 + 5i
= (4 − 5i)/41 → (4 − 5i)/41
(ii) 6 + 2i
= (6 − 2i)/40 = (3 − i)/20
(iii) 7 − 3i
= (7 + 3i)/58
(iv) √5 − 4i
= (√5 + 4i)/21
Q.4 — Verify the Following (z₁ = 2 + 5i, z₂ = 1 − 3i, z₃ = 2 + i)
(i) z₁ + z₂ = z₂ + z₁
Both sides = 3 + 2i ✓ Verified
(ii) z₁z₂ = z₂z₁
Both sides = 17 − i ✓ Verified
(iii) (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Both sides = 5 + 3i ✓ Verified
(iv) (z₁z₂)z₃ = z₁(z₂z₃)
Both sides = 35 + 15i ✓ Verified
(v) z₁ + (−z₁) = (−z₁) + z₁ = 0
Both sides = 0 ✓ Verified
Q.5 — If (1 + i)²/(2 − i) = x + iy, Find x and y
(1 + i)² = 2i, then 2i/(2 − i) × (2 + i)/(2 + i) = (4i − 2)/5
x = −2/5, y = 4/5
Q.6 — If (2x + iy)(1 − i) = 4 + 2i, Find x and y
Expanding gives: 2x + y = 4 and y − 2x = 2
x = 1/2, y = 3
Q.7 — Find a and b, if (a + bi)(1 + 3i) = −8 + 11i
Expanding gives: a − 3b = −8 and 3a + b = 11
a = 5/2, b = 7/2
Key Facts — Exercise 1.2
Powers of i: i⁰ = 1, i¹ = i, i² = −1, i³ = −i, i⁴ = 1 (cycle repeats)
Conjugate & Modulus: if z = a + bi, then z̄ = a − bi, and |z| = √(a² + b²)
Multiplicative inverse: if z = a + bi (z ≠ 0), then z⁻¹ = (a − bi)/(a² + b²)
Division of complex numbers: (a + bi)/(c + di) × (c − di)/(c − di) = [(a+bi)(c−di)] / (c² + d²)
Additive inverse: if z = a + bi, then −z = −a − bi
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Frequently Asked Questions
How do you divide two complex numbers?
Multiply the numerator and denominator by the conjugate of the denominator, then simplify using i² = −1.
What is the multiplicative inverse of a complex number?
For z = a + bi, the multiplicative inverse is (a − bi)/(a² + b²), found by multiplying by the conjugate over itself.
What is the difference between additive inverse and multiplicative inverse?
The additive inverse of z = a + bi is −a − bi (adds to zero), while the multiplicative inverse is 1/z (multiplies to give 1).
What is Exercise 1.2 of Class 10 Maths Chapter 1 about?
It covers the four basic operations on complex numbers (addition, subtraction, multiplication, division), additive and multiplicative inverses, and verifying algebraic properties like commutativity and associativity.
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