Question

$\frac{3+2isin\theta }{1-2isin\theta }$ will be real, if $\theta$ =
[IIT 1976; EAMCET 2002]

• A. $2n\pi$
• B. $n\pi +\frac{\pi }{2}$
• C. $n\pi$
• D. None of these

Answer 1

The correct option is C $n\pi$

$\frac{(3+2isin\theta )(1+2isin\theta )}{(1-2isin\theta )(1+2isin\theta )}=\left(\frac{3-4si{n}^2\theta }{1+4si{n}^2\theta }\right)+i\left(\frac{8sin\theta }{1+4si{n}^2\theta }\right)$
Now, since it is real, therefore Im (z) = 0
$\Rightarrow \frac{8sin\theta }{1+4si{n}^2\theta }=0\Rightarrow sin\theta =0,\therefore \theta =n\pi$

where n = 0, 1, 2, 3, ......

Trick : Check for (a), if the given number is absolutely real but (c) also satisfies this condition and in (a) and (c), (c) is most general value of $\theta$.