De Moivre’s Theorem
- It is customary to write cisθ for cosθ + i sinθ. Thus, we may state the De Moivre’s theorem as (cisθ)ⁿ = cis(nθ) if n ϵ z.
- (cosθ + i sinθ)⁻ⁿ = cos(-nθ) + i sin(-nθ) = cos(nθ) – isin(nθ) provided n is an integer.
- (cosθ + isinθ) (cosθ – isinθ) = cos²θ – i²sin²θ = cos²θ + sin²θ = 1.
- (cosθ +isinθ) =\(\frac{1}{\cos \theta -i\sin \theta }.(\cos \theta -i\sin \theta )=\frac{1}{\cos \theta +i\sin \theta }\).
- (cosθ – isinθ)ⁿ = \({{\left( \frac{1}{\cos \theta +i\sin \theta } \right)}^{n}}\) = (cosθ + isinθ)⁻ⁿ = cosnθ – isinnθ provided n an integer.
- cisθ. cisɸ = cis (θ + ɸ) for any θ, ɸ ϵ R.
Example: simplify \(\frac{{{\left( \cos \alpha +i\sin \alpha \right)}^{4}}}{{{(\sin \beta +i\cos \beta )}^{8}}}\).
Solution: Given that \(\frac{{{\left( \cos \alpha +i\sin \alpha \right)}^{4}}}{{{(\sin \beta +i\cos \beta )}^{8}}}\),
\(\frac{{{\left( \cos \alpha +i\sin \alpha \right)}^{4}}}{{{(\sin \beta +i\cos \beta )}^{8}}}=\frac{{{(\cos \alpha +i\sin \alpha )}^{4}}}{{{(-{{i}^{2}}\sin \beta +i\cos \beta )}^{8}}}\),
\(=\frac{{{(\cos \alpha +i\sin \alpha )}^{4}}}{{{(i)}^{8}}{{(\cos \beta -i\sin \beta )}^{8}}}\),
(cosα + isinα)⁴ (cosβ – isinβ)⁻⁸.
(since i⁸ = 1).
= (cos4α + isin4α) (cos (-8) β – isin (-8 β)).
= (cos4 α + isin4 α) (cos8 β + isin8 β).