COMPLEX NUMBERS · HL
De Moivre's Theorem
Use proof by induction to prove.
Section 1 of 5
The statement
$z = (\cos\theta + i\sin\theta)$ use proof by induction to prove
$z^{n} = (\cos n\theta + i\sin n\theta)$
Prove:
$(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta$
Section 2 of 5
Prove for n = 1
$\cos\theta + i\sin\theta = \cos\theta + i\sin\theta$
Section 3 of 5
Assume for n = k
$(\cos\theta + i\sin\theta)^{k} = \cos k\theta + i\sin k\theta$
Section 4 of 5
Prove n = k + 1
$(\cos\theta + i\sin\theta)^{k+1} = \cos(k+1)\theta + i\sin(k+1)\theta$
$(\cos\theta + i\sin\theta)^{k}(\cos\theta + i\sin\theta)$
$(\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$
$\cos k\theta\cos\theta + i\cos k\theta\sin\theta + i\sin k\theta\cos\theta + i^{2}\sin k\theta\sin\theta$
$\cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\cos k\theta\sin\theta + \sin k\theta\cos\theta)$
$\cos(k\theta + \theta) + i\sin(k\theta + \theta)$
$\cos(k+1)\theta + i\sin(k+1)\theta$
$\cos(A+B) = \cos A\cos B - \sin A\sin B$
$\sin(A+B) = \sin A\cos B + \cos A\sin B$
Section 5 of 5
Conclusion
no matter what is written down
Since true for $n = 1$ and $n = k + 1$ then true for all $n$.
SUM
The lot in one box
De Moivre's Theorem toolkit
1.$(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta$
2.$\cos(A+B) = \cos A\cos B - \sin A\sin B$
3.$\sin(A+B) = \sin A\cos B + \cos A\sin B$
End of lesson
De Moivre's Theorem — HL · Mathslive.ie