Question

Consider the complex numbers $z=\frac{(1-isin\theta )}{(1+icos\theta )}$. The value of $\theta$ for which z is purely imaginary are-

• A. $n\pi -\frac{\pi }{4},nϵI$
• B. $n\pi +\frac{\pi }{4},nϵI$
• C. $n\pi ,nϵI$
• D. No real values of $\theta$

Answer 1

Answer: (D)

No real values of $\theta$

Multiplying and dividing by $1-icos\theta$
$\frac{(1-isin\theta )(1-icos\theta )}{1+co{s}^2\theta }$
$=\frac{1-i(cos\theta +sin\theta )-\frac{sin2\theta }{2}}{1+co{s}^2\left(\theta \right)}$
$=\frac{1-\frac{sin2\theta }{2}}{1+co{s}^2\left(\theta \right)}-i\frac{(cos\theta +sin\theta )}{1+co{s}^2\left(\theta \right)}$
If $z$ is purely imaginary.
Then
$\frac{1-\frac{sin2\theta }{2}}{1+co{s}^2\left(\theta \right)}=0$
$1-\frac{sin2\theta }{2}=0$
$1=\frac{sin2\theta }{2}$
$sin2\theta =2$
This contradictory, since, $sin\theta ϵ[-1,1]$
Hence, no values of $\theta$ is possible.