40
Product and quotient of complex numbers in trigonometric form
Textbook, Part 2: pp. 81–83
GoalMultiply and divide complex numbers in trigonometric form and raise them to a power by De Moivre’s formula.
New words
product · ko‘paytmaquotient · bo‘linmaDe Moivre’s formula · Muavr formulasisum of arguments · argumentlar yig‘indisi
Explanation
If z₁ = r₁(cos φ₁ + i sin φ₁) and z₂ = r₂(cos φ₂ + i sin φ₂), then the product is z₁z₂ = r₁r₂[cos(φ₁ + φ₂) + i sin(φ₁ + φ₂)]: moduli multiply and arguments add. The quotient is z₁/z₂ = (r₁/r₂)[cos(φ₁ − φ₂) + i sin(φ₁ − φ₂)]: moduli divide and arguments subtract. These rules follow from the addition formulas for sine and cosine. Multiplying z = r(cos φ + i sin φ) by itself n times gives De Moivre’s formula: zⁿ = rⁿ(cos nφ + i sin nφ), n ∈ ℕ. For high powers (for example (1 + i)¹⁰) it is easier to go to trigonometric form first than to multiply in algebraic form. If the argument exceeds 2π we drop the multiple of 2π.
Worked examples
z₁ = 2(cos 40° + i sin 40°), z₂ = 3(cos 20° + i sin 20°). z₁z₂ = 6(cos 60° + i sin 60°) = 6(1/2 + (√3/2)i) = 3 + 3√3 i.
(1 + i)⁸: 1 + i = √2(cos 45° + i sin 45°). By De Moivre: (√2)⁸(cos 360° + i sin 360°) = 16 · 1 = 16.
Class activity
In pairs compute (1 + i)², (1 + i)⁴, (1 + i)⁸ successively by the algebraic way and compare with the De Moivre result (16).
Practice
1
What are the modulus and argument of the product of two numbers in trigonometric form?
The product of the moduli and the sum of the arguments.
2
Divide 4(cos 100° + i sin 100°) by 2(cos 40° + i sin 40°). State the modulus of the result.
2
3
Compute (cos 15° + i sin 15°)⁶.
i
4
Why does (cos φ + i sin φ)ⁿ always have modulus 1?
The modulus is 1ⁿ = 1 by De Moivre; equivalently cos² nφ + sin² nφ = 1.