Tamil Nadu HSE Second Year Statistics study materials
Tamil Nadu HSE Second Year Statistics study resources

Tamil Nadu HSE Second Year Statistics: Complex Numbers

Tamil Nadu HSE Second Year Statistics · Work through your exam, one topic at a time.

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Complex Numbers

The Complex Numbers domain tests algebraic manipulation of a + ib, modulus and argument, polar form, De Moivre's theorem, and the roots of unity.

Core rules

For z = a + ib, |z| = √(a² + b²) and z z̄ = |z|². The argument satisfies tan θ = b/a but must be placed in the correct quadrant; the principal argument lies in (−π, π]. In polar form z = r(cos θ + i sin θ), so z₁z₂ has modulus r₁r₂ and argument θ₁ + θ₂, while z₁/z₂ subtracts arguments. De Moivre: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, and the nth roots are r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, …, n − 1.

Cube roots of unity

They are 1, ω = (−1 + i√3)/2 and ω², with 1 + ω + ω² = 0 and ω³ = 1. These identities collapse long powers into a single term.

Worked example

Evaluate (1 + i)⁸. Here r = √2 and θ = π/4, so (1 + i)⁸ = (√2)⁸(cos 2π + i sin 2π) = 16 × 1 = 16. Repeated multiplication is unnecessary once the polar form is written.

Traps

Writing the argument as tan⁻¹(b/a) without adjusting the quadrant; using De Moivre with a non-integer exponent and reporting one answer instead of n; assuming |z₁ + z₂| = |z₁| + |z₂|; forgetting that |z| is real and non-negative.