If sqrt{5}-i sqrt{15}=r(cos theta+i sin theta | vedclass

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Question

(C) Given the complex number $z = \sqrt{5} - i \sqrt{15}$.Comparing with $r(\cos 	heta + i \sin 	heta)$,we have $r = |z| = \sqrt{(\sqrt{5})^2 + (-\s

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Given the complex number $z = \sqrt{5} - i \sqrt{15}$.Comparing with $r(\cos 	heta + i \sin 	heta)$,we have $r = |z| = \sqrt{(\sqrt{5})^2 + (-\sqrt{15})^2} = \sqrt{5 + 15} = \sqrt{20} = 2 \sqrt{5}$.Thus,$r^2 = 20$.We have $\cos 	heta = \frac{\sqrt{5}}{2 \sqrt{5}} = \frac{1}{2}$ and $\sin 	heta = \frac{-\sqrt{15}}{2 \sqrt{5}} = -\frac{\sqrt{3}}{2}$.Then $\sec 	heta = \frac{1}{\cos 	heta} = 2$.And $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta} = -\frac{2}{\sqrt{3}}$,so $\operatorname{cosec}^2 	heta = \frac{4}{3}$.Substituting these values into the expression $r^2(\sec 	heta + 3 \operatorname{cosec}^2 	heta)$:$20 	imes (2 + 3 	imes \frac{4}{3}) = 20 	imes (2 + 4) = 20 	imes 6 = 120$.

If $\sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta)$ where $-\pi < \theta < \pi$,then find the value of $r^2(\sec \theta+3 \operatorname{cosec}^2 \theta)$.

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