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Find real $\;\theta\;$ such that $\;\large\frac{3 +2i sin \theta}{1-2isin\theta}\;$ is purely real .

$(a)\;0\qquad(b)\;\pi\qquad(c)\;n\pi \;,n \in Z\qquad(d)\;i \pi$

1 Answer

Answer : $\;n \pi , n \in Z$
Explanation :
$\;\large\frac{3 +2i sin \theta}{1-2isin\theta}\;= \large\frac{3 +2i sin \theta}{1-2isin\theta} \times \large\frac{1+2i sin \theta}{1+2isin \theta}$
$ = \large\frac{3 + 6 i sin \theta + 2i sin \theta + 4i^{2} sin^{2} \theta} {1^{2} - (i sin \theta)^{2}}$
$= \large\frac{3 - 4 sin^{2} \theta}{1+ 4 sin^{2} \theta} + \large\frac{8i sin \theta}{ 1+4 sin^{2} \theta} $
We are given the complex number to be real , therefore
$ \large\frac{8 sin \theta}{ 1+4 sin^{2} \theta} =0$
i.e $\; sim \theta =0$
Then , $n \pi , n \in Z$
answered Apr 16, 2014 by yamini.v
 
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