# De Moivre’s Theorem

## 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).