Question

$\frac{(\cos \theta +i\sin \theta {)}^4}{(\sin \theta +i\cos \theta {)}^5}$ is equal to

• A. $-\cos 9\theta -i\sin 9\theta$
• B. $\cos 9\theta +i\sin 9\theta$
• C. $\sin 9\theta -i\cos 9\theta$
• D. $\sin 9\theta +i\cos 9\theta$

Answer 1

Answer: (C)

$\sin 9\theta -i\cos 9\theta$

$\frac{(\cos \theta +i\sin \theta {)}^4}{(\sin \theta +i\cos \theta {)}^5}=\frac{(\cos \theta +i\sin \theta {)}^4}{i^5(\cos \theta -i\sin \theta {)}^5}$ $[\because i^2=-1)$

$=\frac{(\cos \theta +i\sin \theta {)}^4}{i(\cos \theta -i\sin \theta {)}^5}$

$=-i(\cos \theta +i\sin \theta {)}^4\times {\left(\frac{1}{\cos \theta -i\sin \theta }\right)}^5$

$=-i(\cos \theta +i\sin \theta {)}^4\times (\cos \theta +i\sin \theta {)}^5$ $(\because \frac{1}{\cos \theta +i\sin \theta }=\cos \theta -i\sin \theta )$

$=-i(\cos \theta +i\sin \theta {)}^9$

$=-i(\cos 9\theta +i\sin 9\theta )$ $(\because (\cos \theta +i\sin \theta {)}^m=\cos m\theta +i\sin m\theta )$

$=-i\cos 9\theta +\sin 9\theta$

$\therefore \frac{(\cos \theta +i\sin \theta {)}^4}{(\sin \theta +i\cos \theta {)}^5}=\sin 9\theta -i\cos 9\theta$