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Find real ?, such that \(\rm \frac{5 + 4i cos \theta }{1 - 4i cos\theta}\) is purely real.

A.
? = n?, \(n \rm \epsilon Z\)
B.
? = 2n?, \(n \rm \epsilon Z\)
C.
? = (2n + 1) \(\rm \frac{\pi }{2}, n \rm \epsilon Z\)
D.
None of these

Solution:

Concept:z = a + ib, is a complex number,then, a is called the real part and b is an imaginary part.Calculation:Given= \( \rm \frac{5 + 4i cos ? }{1 - 4i cos?}\)= \(\rm \frac{(5 + 4i cos ?)(1 + 4i cos?) }{(1 - 4i cos?)(1 + 4i cos?)}\)= \(\rm \frac{(5 + 24i cos ? - 16cos^{2}?)}{(1 + 16 cos^{2}?)}\)= \(\rm \frac{5 - 16cos^{2}?}{1 + 16cos^{2}?} + {\frac{24i cos?}{1 + 16cos^{2}?}}\)Complex number to be purely real\(\rm {\frac{24 cos?}{1 + 16cos^{2}?}}\) = 0,cos ? = 0Thus, ? = (2n + 1) \(\rm \frac{\pi }{2}, n \rm \epsilon Z\)

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