CAIE A-Level Further Math A2 2.5.3 De Moivres Theorem Questions

$\frac{z^{n}-1}{z-1}$
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$\frac{z^{n}-1}{z-1}=\frac{\cos n \theta-1+\mathrm{i} \sin n \theta}{\cos \theta-1+\mathrm{i} \sin \theta}$$\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)}$
Multiplies numerator and denominator by complex
conjugate.

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

8(b)

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

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

$=\frac{\sin n \theta \cos \frac{1}{2} \theta}{2 \sin \frac{1}{2} \theta}-\frac{1}{2} \cos n \theta+\frac{1}{2}$
Divides through by denominator.

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

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