CAIE A-Level Further Math 2.5.3 de Moivre's theorem

$\frac{z^{n}-1}{z-1}$

B1

1

$\frac{z^{n}-1}{z-1}=\frac{\cos n \theta-1+\mathrm{i} \sin n \theta}{\cos \theta-1+\mathrm{i} \sin \theta}$

B1

$\frac{(\cos n \theta-1+\mathrm{i} \sin n \theta)(\cos \theta-1-\mathrm{i} \sin \theta)}{(\cos \theta-1+\mathrm{i} \sin \theta)(\cos \theta-1-\mathrm{i} \sin \theta)}$

M1

Multiplies numerator and denominator by complex
conjugate.

$\operatorname{Re}\left(\frac{z^{n}-1}{z-1}\right)=\frac{\cos n \theta \cos \theta+\sin n \theta \sin \theta-\cos n \theta-\cos \theta+1}{(\cos \theta-1)^{2}+\sin ^{2} \theta}$

M1

Takes real part.
$\cos (n-1) \theta=\cos n \theta \cos \theta+\sin n \theta \sin \theta$$=\frac{\cos n \theta \cos \theta+\sin n \theta \sin \theta-\cos n \theta}{2(1-\cos \theta)}+\frac{1}{2}$

A1

$=\frac{\cos n \theta(\cos \theta-1)+\sin n \theta \sin \theta}{2(1-\cos \theta)}+\frac{1}{2}$

M1

Factorises.

8(b)

$=\frac{1}{2}\left(1-\cos n \theta+\frac{\sin n \theta \sin \theta}{1-\cos \theta}\right)$

M1 A1

Divides through by denominator. AG.

Alternative method for question 8(b)

$\frac{z^{n}-1}{z-1}=\frac{\mathrm{e}^{\mathrm{i} n \theta}-1}{\mathrm{e}^{\mathrm{i} \theta}-1}$

B1

$\frac{\mathrm{e}^{\mathrm{i}\left(n-\frac{1}{2}\right) \theta}-\mathrm{e}^{-\mathrm{i} \frac{1}{2} \theta}}{\mathrm{e}^{\mathrm{i} \frac{1}{2} \theta}-\mathrm{e}^{-\mathrm{i} \frac{1}{2} \theta}}=\frac{\cos \left(n-\frac{1}{2}\right) \theta+\mathrm{i} \sin \left(n-\frac{1}{2}\right) \theta-\cos \frac{1}{2} \theta+\mathrm{i} \sin \left(\frac{1}{2} \theta\right)}{2 \mathrm{i} \sin \frac{1}{2} \theta}$

M1

$\operatorname{Re}\left(\frac{z^{n}-1}{z-1}\right)=\frac{\sin \left(n-\frac{1}{2}\right) \theta+\sin \frac{1}{2} \theta}{2 \sin \frac{1}{2} \theta}$

M1

Takes real part

$=\frac{\sin \left(n-\frac{1}{2}\right) \theta}{2 \sin \frac{1}{2} \theta}+\frac{1}{2}$

A1

$=\frac{\sin n \theta \cos \frac{1}{2} \theta-\cos n \theta \sin \frac{1}{2} \theta}{2 \sin \frac{1}{2} \theta}+\frac{1}{2}$

M1

Uses compound angle identity

$=\frac{\sin n \theta \cos \frac{1}{2} \theta}{2 \sin \frac{1}{2} \theta}-\frac{1}{2} \cos n \theta+\frac{1}{2}$

M1

Divides through by denominator.

$\frac{\sin n \theta \sin \theta}{4 \sin ^{2} \frac{1}{2} \theta}-\frac{1}{2} \cos n \theta+\frac{1}{2}=\frac{\sin n \theta \sin \theta}{2(1-\cos \theta)}-\frac{1}{2} \cos n \theta+\frac{1}{2}$

A1

AG. $\sin \theta=2 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta$ and
$2 \sin ^{2} \frac{1}{2} \theta=1-\cos \theta$.

7