## - De Moivre's Theorem

$$z=r(\cos ⁡θ+i \sin ⁡θ)$$ then

$$z^2=r^2 (\cos^2⁡θ-\sin^2⁡θ+2i \sin⁡θ \cos⁡θ)$$
$$=r^2(\cos⁡2θ+i \sin⁡2θ)$$
This can be extended to give
$$z^n=r^n (\cos ⁡nθ+i \sin ⁡nθ)$$
Assuming $$r=1$$ we now have
$$(\cos⁡θ+i \sin⁡θ)^n=\cos ⁡nθ+i \sin ⁡nθ$$

LESSON 1
Use De Moivre’s Theorem to prove the following identities
$$\cos ⁡4θ≡8 \cos^4⁡θ-8 \cos^2⁡θ+1$$
SOLUTION
When $$n=4$$ and $$r=1$$
$$\cos ⁡4θ+i \sin⁡4θ=(\cos⁡θ+i \sin⁡θ)^4$$
• Use BINOMIAL EXPANSION or PASCAL’S TRIANGLE to expand the brackets
$$⟹\cos^4⁡θ+4 \cos^3⁡θ i \sin⁡θ+6 \cos^2⁡θ i^2 \sin^2⁡θ+4 \cos⁡θ i^3 \sin^3⁡θ+i^4 \sin^4⁡θ$$
• Simplify the powers of $$i$$
$$⟹\cos^4⁡θ+4i \cos^3⁡θ \sin⁡θ-6 \cos^2⁡θ \sin^2⁡θ-4i \cos⁡θ \sin^3⁡θ+\sin^4⁡θ$$
• Equate real parts and simplify
$$\cos⁡4θ=\cos^4⁡θ-6 \cos^2\theta \sin^2⁡θ +\sin^4⁡θ$$
$$⟹\cos^4⁡θ-6 \cos^2⁡θ (1-\cos^2⁡θ)+(1-\cos^2⁡θ)(1-\cos^2⁡θ)$$
$$⟹\cos^4⁡θ-6 \cos^2⁡θ+6 \cos^4⁡θ+\cos^4⁡θ-2 \cos^2⁡θ+1$$
$$⟹8 \cos^4⁡θ-8 \cos^2⁡θ+1$$

LESSON 2
Use de Moivre’s theorem to show that
$$\sin⁡5θ=a \cos^4⁡θ \sin⁡θ+b \cos^2⁡θ \sin^3⁡θ+c \sin^5⁡θ$$
where $$a, b$$ and $$c$$ are integers determined.
SOLUTION
When $$n=5$$
$$\cos⁡5θ+i \sin⁡5θ=(\cos⁡θ+i \sin⁡θ)^5$$
• Use BINOMIAL EXPANSION or PASCAL’S TRIANGLE to determine the coefficients for the expansion.
$$⟹1 \cos^5⁡θ+5 \cos^4⁡θ (i \sin⁡θ )+10 \cos^3⁡θ (i \sin⁡θ)^2+10\cos^2⁡θ(i \sin⁡θ )^3+5\cos⁡θ (i \sin⁡θ)^4+1(i \sin⁡θ)^5$$
• Simplify powers of $$i$$
$$⟹\cos^5⁡θ+5i \cos^4⁡θ \sin⁡θ-10 \cos^3⁡θ \sin^2⁡θ-10i \cos^2⁡θ \sin^3⁡θ+5 \cos⁡θ \sin^4⁡θ+i \sin^5⁡θ$$
• Equate Imaginary Parts
$$\sin⁡5θ=5 \cos^4⁡θ \sin⁡θ-10 \cos^2⁡θ \sin^3⁡θ+\sin^5⁡θ$$
$$a=5, b=-10, c=1$$

LESSON 3
Use de Moivre’s theorem to show that
$$\displaystyle \tan⁡3θ={3 \tan⁡θ-\tan^3⁡θ\over 1-3 \tan^2⁡θ}$$
SOLUTION
$$\cos⁡3θ+i \sin⁡3θ=(\cos⁡θ+i \sin⁡θ)^3$$
$$=\cos^3⁡θ+3 \cos^2⁡θ (i \sin⁡θ )+3 \cos⁡θ (i \sin⁡θ)^2+(i \sin⁡θ )^3$$
$$=\cos^3⁡θ+3i \cos^2⁡θ \sin⁡θ-3 \cos⁡θ \sin^2⁡θ-i \sin^3⁡θ$$
• Equate Real Parts:
$$\cos⁡3θ=\cos^3⁡θ-3 \cos⁡θ \sin^2⁡θ$$
• Equate Imaginary Parts
$$\sin⁡3θ=3 \cos^2⁡θ \sin⁡θ-\sin^3⁡θ$$
• Use the result
$$\displaystyle \tan⁡3θ={\sin⁡3θ\over \cos⁡3θ}$$
$$\displaystyle ⟹{3 \cos^2⁡θ \sin⁡θ-\sin^3⁡θ\over \cos^3⁡θ-3 \cos⁡θ \sin^2⁡θ}$$
• Divide throughout by the highest power of $$\cos⁡θ$$
$$\displaystyle ⟹{{3 \cos^2⁡θ \sin⁡θ\over \cos^3⁡θ} -{\sin^3⁡θ\over \cos^3⁡θ}\over {\cos^3⁡θ\over \cos^3⁡θ} -{3 \cos⁡θ \sin^2⁡θ\over \cos^3⁡θ}}$$
$$\displaystyle⟹{3 \tan⁡θ-\tan^3⁡θ\over 1-3 \tan^2⁡θ}$$

LESSON 4
Find the value of
(i) $$\left(\cos⁡\left({π\over4}\right)+i \sin⁡\left({π\over4}\right)\right)^{12}$$
(ii) $$\displaystyle {1\over (1-i\sqrt{3}})^3$$
SOLUTION
(i) $$\left(\cos \left({\pi \over 4}\right)+i \sin\left({\pi\over4}\right)\right)^{12}$$
By De Moivre’s Theorem
$$\left(\cos \left({\pi \over 4}\right)+i \sin\left({\pi\over4}\right)\right)^{12}=\cos\left(12\times {\pi\over4}\right)+i\sin\left(12\times{\pi\over4}\right)$$
$$=\cos⁡3π+i \sin⁡3π$$
$$=-1$$
(ii) $$\displaystyle {1\over (1-i\sqrt{3}})^3$$
• Rewrite $$1-i\sqrt{3}$$ in Modulus – Argument form
$$|1-\sqrt{3} i|=\sqrt{1^2+(-\sqrt{3})^2}=2$$
$$\displaystyle θ=-\tan^{-1}⁡(\sqrt{3}$$
$$\displaystyle =-{π\over3}$$
By De Moivre’s Theorem
$$\displaystyle (1-\sqrt{3} i)^{-3}=\left[2\left(\cos\left(-{\pi\over3}\right)+i\sin\left(-{\pi\over3}\right)\right)\right]^{-3}$$
$$\displaystyle =2^{-3}\left(\cos⁡\left(-3\left(-{π\over3}\right)\right)+i \sin⁡\left(-3\left(-{π\over3}\right)\right)\right)$$
$$\displaystyle ={1\over8}(\cos⁡π+i \sin⁡π)$$
$$\displaystyle ={1\over8} (-1)$$
$$\displaystyle =-{1\over8}$$

LESSON 5
Express $$\sqrt{3}+i$$ in the modulus – argument form. Hence, find $$(\sqrt{3}+i)^{10}$$ in the form $$a+bi$$.
SOLUTION
$$|\sqrt{3}+i|=\sqrt{(\sqrt{3})^2+1^2}=2$$
$$\displaystyle \theta=\tan^{-1}⁡\left({1\over \sqrt{3}}\right)={π\over6}$$
$$\displaystyle \sqrt{3}+i=2\left(\cos\left({\pi\over6}\right)+i\sin\left({\pi\over6}\right)\right)$$
By De Moivre’s Theorem
$$\displaystyle (\sqrt{3}+i)^{10}=\left[2\left(\cos\left({π\over6}\right)\right)+i\sin\left({π\over6}\right)\right]^{10}$$
$$\displaystyle (\sqrt{3}+i)^{10}=2^{10}\left[\left(\cos\left(10\left({π\over6}\right)\right)\right)+i\sin\left(10\left({π\over6}\right)\right)\right]^{10}$$
$$\displaystyle ⟹1024\left(\cos\left({5π\over3}\right)+i\sin\left({5π\over3}\right)\right)$$
$$\displaystyle ⟹1024\left({1\over2}-{\sqrt{3}\over 2} i\right)$$
$$⟹512-512\sqrt{3} i$$