Question

Consider the complex numbers $z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )}$ and if argument of $z$ is $\frac{\pi }{4}$, then

• A. $\theta =n\pi ,n\in I$ 0nly
• B. $\theta =(2n+1),n\in I$ only.
• C. both $\theta =n\pi$ and $\theta =(2n+1)\frac{\pi }{2}$, $nϵI$
• D. none of these

Answer 1

Answer: (D)

none of these

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 $arg\left(z\right)$ is $\frac{\pi }{4}$, then

$1-\frac{sin2\theta }{2}=-(cos\theta +sin\theta )$

$\frac{2-2sin\theta cos\theta }{2}=-(cos\theta +sin\theta )$

$1+(cos\theta -sin\theta {)}^2=-2(cos\theta +sin\theta )$

$\theta =(2n-1)\pi$ satisfies, the above equation.