Inter 2nd Year Maths 2A De Moivre’s Theorem Formulas

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 2 De Moivre’s Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Formulas

Statement:
→ If ‘n’ is an integer, then (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
If n’ is a rational number, then one of the values of
(cos θ + i sin θ)ⁿ is cos nθ + i sin nθ

n^th roots of unity:
→ n^th roots of unity are {1, ω, ω² …….. ω^{n – 1}}.
Where ω = \(\left[\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right]\) k = 0, 1, 2 ……. (n – 1).
If ω is a n^th root of unity, then

• ωⁿ = 1
• 1 + ω + ω² + ………… + ω^{n – 1} = 0

Cube roots of unity:
→ 1, ω, ω² are cube roots of unity when

• ω³ = 1
• 1 + ω + ω² = 0
• ω = \(\frac{-1+i \sqrt{3}}{2}\), ω² = \(\frac{-1-i \sqrt{3}}{2}\)
• Fourth roots of unity roots are 1, – 1, i, – i

→ If Z₀ = r₀ cis θ₀ ≠ 0, then the n^th roots of Z₀ are α_k = (r₀)^1/n cis\(\left(\frac{2 k \pi+\theta_{0}}{n}\right)\) where k = 0, 1, 2, ……… (n – 1)

→ If n is any integer, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

→ If n is any fraction, one of the values of (cosθ + i sinθ)ⁿ is cos nθ + i sin nθ.

→ (sinθ + i cosθ)ⁿ = cos(\(\frac{n \pi}{2}\) – nθ) + i sin(\(\frac{n \pi}{2}\) – nθ)

→ If x = cosθ + i sinθ, then x + \(\frac{1}{x}\) = 2 cosθ, x – \(\frac{1}{x}\) = 2i sinθ

→ xⁿ + \(\frac{1}{x^{n}}\) = 2cos nθ, xⁿ – \(\frac{1}{x^{n}}\) = 2i sin nθ

→ The nth roots of a complex number form a G.P. with common ratio cis\(\frac{2 \pi}{n}\) which is denoted by ω.

→ The points representing n^th roots of a complex number in the Argand diagram are concyclic.

→ The points representing n^th roots of a complex number in the Argand diagram form a regular polygon of n sides.

→ The points representing the cube roots of a complex number in the Argand diagram form an equilateral triangle.

→ The points representing the fourth roots of complex number in the Argand diagram form a square.

→ The nth roots of unity are 1, w, w²,………. , w^n-1 where w = cis\(\frac{2 \pi}{n}\)

→ The sum of the nth roots of unity is zero (or) the sum of the nth roots of any complex number is zero.

→ The cube roots of unity are 1, ω, ω² where ω = cis\(\frac{2 \pi}{3}\), ω² = cis\(\frac{4 \pi}{3}\) or
ω = \(\frac{-1+i \sqrt{3}}{2}\)
ω² = \(\frac{-1-i \sqrt{3}}{2}\)
1 + ω + ω² = 0
ω³ = 1

→ The product of the n^th roots of unity is (-1)^n-1 .

→ The product of the n^th roots of a complex number Z is Z(-1)^n-1 .

→ ω, ω² are the roots of the equation x² + x + 1 = 0