- 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
\(\displaystyle (\sqrt{3}+i)^{10}=\left[2\left(\cos\left({π\over6}\right)\right)+i\sin\left({π\over6}\right)\right]^{10}\)
\(\displaystyle (\sqrt{3}+i)^{10}=2^{10}\left[\left(\cos\left(10\left({π\over6}\right)\right)\right)+i\sin\left(10\left({π\over6}\right)\right)\right]^{10}\)
\(\displaystyle ⟹1024\left(\cos\left({5π\over3}\right)+i\sin\left({5π\over3}\right)\right)\)
\(\displaystyle ⟹1024\left({1\over2}-{\sqrt{3}\over 2} i\right)\)
\(⟹512-512\sqrt{3} i\)