Complex Numbers And Quadratic Equations question 413
Question: $ \frac{3+2i\sin \theta }{1-2i\sin \theta } $ will be real, if $ \theta $ = [IIT 1976; EAMCET 2002]
Options:
A) $ 2n\pi $
B) $ n\pi +\frac{\pi }{2} $
C) $ n\pi $
D) None of these [Where $ n $ is an integer]
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{(3+2i\sin \theta )(1+2i\sin \theta )}{(1-2i\sin \theta )(1+2i\sin \theta )} $ = $ ( \frac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta } )+i( \frac{8\sin \theta }{1+4{{\sin }^{2}}\theta } ) $ Now, since it is real, therefore Im $ (z)=0 $
Þ $ \frac{8\sin \theta }{1+4{{\sin }^{2}}\theta } $ = 0
Þ $ \sin \theta =0 $ ,
$ \therefore $ $ \theta =n\pi $ where $ n=0 $ , 1, 2, 3, …… Trick: Check for (a), if $ n=0,\theta =0 $ the given number is absolutely real but (c) also satisfies this condition and in (a) and (c), (c) is most general value of $ \theta $ .
BETA
