If z = cos alpha + i sin alpha; 0

Question

(A) Given $z = \cos \alpha + i \sin \alpha$,where $0 < \alpha < \frac{\pi}{4}$.Using De Moivre's Theorem,$z^n = \cos(n\alpha) + i \sin(n\alpha)$.$\fra

Vedclass · Accepted Answer

$\frac{\cos 2 \alpha}{\sin \frac{3}{2} \alpha}$

Vedclass · Answer

(A) Given $z = \cos \alpha + i \sin \alpha$,where $0 Using De Moivre's Theorem,$z^n = \cos(n\alpha) + i \sin(n\alpha)$.$\frac{1+z^4}{1-z^3} = \frac{1 + \cos 4\alpha + i \sin 4\alpha}{1 - \cos 3\alpha - i \sin 3\alpha}$$= \frac{2 \cos^2 2\alpha + 2i \sin 2\alpha \cos 2\alpha}{2 \sin^2 \frac{3\alpha}{2} - 2i \sin \frac{3\alpha}{2} \cos \frac{3\alpha}{2}}$$= \frac{2 \cos 2\alpha (\cos 2\alpha + i \sin 2\alpha)}{2 \sin \frac{3\alpha}{2} (\sin \frac{3\alpha}{2} - i \cos \frac{3\alpha}{2})}$Taking the modulus on both sides:$\left|\frac{1+z^4}{1-z^3}ight| = \left| \frac{\cos 2\alpha}{\sin \frac{3\alpha}{2}} ight| 	imes \frac{|\cos 2\alpha + i \sin 2\alpha|}{|\sin \frac{3\alpha}{2} - i \cos \frac{3\alpha}{2}|}$Since $|\cos 	heta + i \sin 	heta| = 1$ and $|\sin 	heta - i \cos 	heta| = | -i(\cos 	heta + i \sin 	heta) | = |-i| 	imes 1 = 1$,$\left|\frac{1+z^4}{1-z^3}ight| = \frac{\cos 2\alpha}{\sin \frac{3\alpha}{2}}$.

If $z = \cos \alpha + i \sin \alpha$; $0 < \alpha < \frac{\pi}{4}$,then $\left|\frac{1+z^4}{1-z^3}\right| = $

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