# Applications of De-Moivre’s Theorem

De-Moivre’s Theorem: A simple formula for calculating powers of complex numbers in the form of cosθ and sinθ is known as de-Moivre’s theorem. If n is a rational number, then (cosθ + sinθ)ⁿ = cos nθ + i sin nθ.

Applications of De-Moivre’s Theorem:

• If z = (cosθ₁ + i sinθ₁) (cosθ₂ + i sinθ₂) … (cosθn + i sinθn) then, z = cos (θ₁ + θ₂ + … + θn) + i sin (θ₁ + θ₂ + … + θn).
• If z = r (cosθ + i sinθ) and n is a positive integer, then $${{(z)}^{1/n}}={{r}^{1/n}}\left[ \cos \left( \frac{2k\pi +\theta }{n} \right)+i\sin \left( \frac{2k\pi +\theta }{n} \right) \right]$$. Where, k = 0, 1, 2, 3, … (n – 1).
• (cosθ – i sinθ)ⁿ = cos nθ – i sin nθ
• 1/ (cosθ + i sinθ) = (cosθ + i sinθ)⁻¹ = cosθ – i sinθ
• (sinθ ± i cosθ)ⁿ ≠ sin nθ ± i cos nθ
• $${{\left( \sin \theta +i\cos \theta \right)}^{n}}={{\left[ \cos \left( \frac{\pi }{2}-\theta \right)+i\sin \left( \frac{\pi }{2}-\theta \right) \right]}^{n}}$$$$=\left[ \cos \left( \frac{n\pi }{2}-n\theta \right) \right]+i\left[ \sin \left( \frac{n\pi }{2}-n\theta \right) \right]$$.
• (cosθ + i sin ø)ⁿ = cos nθ ± i sin nø

1. If a = cosθ + i sinθ, then the value of $$\frac{1+a}{1-a}$$ is

Solution: a = cosθ + i sinθ

∴ $$\frac{1+a}{1-a}=\frac{1+\cos \theta +i\sin \theta }{1-(\cos \theta +i\sin \theta )}$$,

$$=\frac{\left( 2{{\cos }^{2}}\frac{\theta }{2} \right)+i\left( 2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right)}{\left( 1-\cos \theta \right)-i\sin \theta }\left( \because 1+\cos \theta =2{{\cos }^{2}}\frac{\theta }{2} \right)$$,

$$=\frac{2{{\cos }^{2}}\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}-2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\left( \because 1-\sin \theta =2{{\sin }^{2}}\frac{\theta }{2} \right)$$.

On multiplying by I in numerator and denominator,

$$=\frac{2\cos \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}{2\sin \frac{\theta }{2}\left( \sin \frac{\theta }{2}-i\cos \frac{\theta }{2} \right)}=\frac{i\cot \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}{\left( i\sin \frac{\theta }{2}-{{i}^{2}}\cos \frac{\theta }{2} \right)}$$,

$$=\frac{i\cot \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}{\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}=-i\cot \frac{\theta }{2}$$.

If $${{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right)$$, then the value of x₁.x₂.x₃ … ∞ is

∵ $${{x}_{r}}\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right)$$.

∴ $${{x}_{1}}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$$.

$${{x}_{2}}=\cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}}$$

…   …   …   …

…   …   …   …

∴ $${{x}_{1}}\,\,{{x}_{2}}\,\,{{x}_{3}}\,…=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right)\,…$$,

$$=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\,… \right)+i\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\,… \right)$$,

$$=\cos \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)+i\sin \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)$$,

= cosπ + i sinπ = -1.