Complex Numbers
- The value of \((1+i)^8\) is Answer: C. 16 (1+i)²=2i⇒(1+i)⁸=(2i)⁴=16 . 1
- If \(\omega\) is a complex cube root of unity, then \((1+\omega-\omega^2)^3=\) Answer: D. −8 1+ω=-ω²⇒(-2ω²)³=-8ω⁶=-8 . 1
- z – z¯ = 6i ( z – z¯ = 6i) z – z¯ = (2 + 3i) – (2 – 3i) = 2 + 3i – 2 + 3i = 6i 1
- (1 + 2i) 2 (1 – i) Let z = (1 + 2i)² (1 – i) = (1 + 4i + 4i²) (1 – i) = [1 + 4i + 4(-1)] (1 – i) ….[∵ i² = -1] = (-3 + 4i) (1 – i) = -3 + 3i + 4i – 4i² = -3 +... 1
- (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) ω is the complex cube root of unity. ω³ = 1 and 1 + ω + ω² = 0 Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1 (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸... 1
- (1 – ω – ω2)3 + (1 – ω + ω2)3 ω is the complex cube root of unity. ω³ = 1 and 1 + ω + ω² = 0 Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1 (1 – ω – ω²)³ + (1 – ω + ω²)³... 1
- (1 + ω – ω2) 6 = 64 Proof: since 1 + ω = -ω² , the expression becomes (-2ω²)⁶ = 64 . Here ω is a complex cube root of unity, so ω³ = 1 and 1 + ω + ω² = 0 , whic... 1
- If ω is the complex cube root of unity, show that (i) (2 – ω)(2 – ω2) = 7 ω is the complex cube root of unity. ω³ = 1 and 1 + ω + ω² = 0 Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1 L.H.S. = (2 – ω)(2 – ω²) = 4 –... 1
- If ω is a complex cube root of unity, prove that (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128. ω is the complex cube root of unity. ∴ ω³ = 1 and 1 + ω + ω² = 0 Also, 1 + ω² = -ω, 1 + ω = -ω² ∴ L.H.S. = (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = [... 1
- 18i complex number Let √18i = a + bi, where a, b ∈ R. Squaring on both sides, we get 18i = a² + b² i² + 2abi 0 + 18i = a² – b² + 2abi …..[∵ i² = -1] Equating r... 1
- i131 + i49 complex number i¹³¹ + i⁴⁹ = (i⁴)³² . i³ + (i⁴)¹² . i = (1)³² (-i) + (1)¹² . i = -i + i = 0 2
- 2x + i9 y(2 + i) = x i7 + 10 i16 2x + i⁹ y(2 + i) = x i⁷ + 10 i¹⁶ 2x + (i⁴)² . i . y(2 + i) = x(i²)³ . i + 10 . (i⁴)⁴ 2x + (1)² . iy(2 + i) = x(-1)³ . i + 10(1)⁴ ……..[∵ i² =... 1