Review Exercise 1 Class 10 Maths Chapter 1 (Complex Numbers) — complete step-by-step solutions for the new book. Free PDF download, Al Huda Science Academy.
Review Exercise 1 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 Review Exercise 1, Chapter 1 (Complex Numbers) for Class 10 Mathematics — New Book. It brings together everything from the chapter — conjugate identities, modulus, finding real/imaginary parts, and solving simultaneous equations with complex coefficients — in one final review set.
Q.5 — If z₁ = 3 + 4i and z₂ = 2 + 3i, Verify
(i) conjugate(z₁ + z₂) = conjugate(z₁) + conjugate(z₂)
Both sides = 5 − 7i ✓ Verified
(ii) conjugate(z₁z₂) = conjugate(z₁)·conjugate(z₂)
Both sides = −6 − 17i ✓ Verified
(iii) conjugate(z₁/z₂) = conjugate(z₁)/conjugate(z₂)
Both sides = 18/13 + (1/13)i ✓ Verified
(iv) |z₁| = |−z₁|
Both equal 5 ✓ Verified
(v) conjugate(conjugate(z₂)) = z₂
2 + 3i = 2 + 3i ✓ Verified
(vi) z₁·conjugate(z₁) = |z₁|²
Both equal 25 ✓ Verified
Q.6 — If z₁ = 5 + 4i and z₂ = 3 + 2i, Find
(i) z₁z₂
= 15 + 22i − 8 = 7 + 22i
(ii) z₁/z₂
= (23 + 2i)/13 = 23/13 + (2/13)i
(iii) conjugate(z₁)·conjugate(z₂)
= 15 − 22i − 8 = 7 − 22i
(iv) |conjugate(z₁)·conjugate(z₂)|
= |z₁|·|z₂| = √41 × √13 = √533
Q.7 — Find the Real and Imaginary Parts of z = (2 + 7i)⁻¹
Multiply by the conjugate (2 − 7i): z = (2 − 7i)/53
Real Part = 2/53, Imaginary Part = −7/53
Q.8 — Solve the Simultaneous Equations for z and w
iz + (2 − i)w = 4 + i …(1) iz + (3 + i)w = 3 + 3i …(2)
Step 1 — Find w: Subtracting (1) from (2): (1 + 2i)w = −1 + 2i → w = (3 + 4i)/5 = 3/5 + (4/5)i
Step 2 — Find z: Substituting w into equation (1) gives iz = 2, so z = 2/i = −2i
z = −2i, w = 3/5 + (4/5)i
Q.9 — Solve (3 − 4i)(a + bi) = 1 + 0i for a and b
Expanding and equating real and imaginary parts gives: 3a + 4b = 1 and 3b − 4a = 0
Solving simultaneously: a = 3/25, b = 4/25
Q.10 — Solve (3 − 2i)(x + yi) = 2(x − 2yi) + 2i − 1 for x and y
Expanding both sides and equating real and imaginary parts gives: x + 2y = −1 and 7y − 2x = 2
Solving simultaneously: x = −1, y = 0
Key Facts — Review Exercise 1 Summary
Conjugate rule: conjugate of a + bi = a − bi
Fundamental identity: i² = −1
Modulus identities:
- z × conjugate(z) = |z|²
- |a + bi| = √(a² + b²)
Double conjugate: conjugate(conjugate(z)) = z
Conjugate distributes over operations:
- conjugate(z₁ + z₂) = conjugate(z₁) + conjugate(z₂)
- conjugate(z₁z₂) = conjugate(z₁)·conjugate(z₂)
- conjugate(z₁/z₂) = conjugate(z₁)/conjugate(z₂), for z₂ ≠ 0
Tip: Always multiply by the conjugate of the denominator to perform division of complex numbers.
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Frequently Asked Questions
What topics does Review Exercise 1 of Chapter 1 (Complex Numbers) cover?
It’s a mixed review covering conjugate identities, modulus, multiplication and division of complex numbers, finding real/imaginary parts of inverses, and solving simultaneous equations with complex coefficients.
What is the easiest way to solve simultaneous equations with complex coefficients?
Use elimination or substitution just like with real numbers, then simplify using i² = −1 whenever two complex terms are multiplied, and rationalize any complex denominator by multiplying with its conjugate.
How do you verify that a conjugate identity holds for two complex numbers?
Calculate both sides of the identity independently — the left-hand side using the operation first then taking the conjugate, and the right-hand side using the conjugates first then the operation — and confirm they match.
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