Question

The imaginary part of $(z-1)(\cos \alpha -i\sin \alpha )+(z-1{)}^{-1}\times (\cos \alpha +i\sin \alpha )$ is zero, if

• A. $|z-1|=1$
• B. $\arg (z-1)=2\alpha$
• C. $\arg (z-1)=\alpha$
• D. $|z-1|=2$

Answer 1

Answer: (A), (C)

$\arg (z-1)=\alpha$

Let $z-1=r(\cos \theta +i\sin \theta )=r{e}^{i\theta }$
$\therefore$ Given expression
$=r{e}^{i\theta }\cdot e^{-i\alpha }+\frac{1}{r{e}^{i\theta }}{e}^{i\alpha }$
$=r{e}^{i(\theta -\alpha )}+\frac{1}{r}{e}^{-i(\theta -\alpha )}$
Since, imaginary part of given expression is zero, we have
$r\sin (\theta -\alpha )-\frac{1}{r}\sin (\theta -\alpha )=0\Rightarrow \sin (\theta -\alpha )(r-\frac{1}{r})=0\Rightarrow r^2=1$
$\Rightarrow r=1$
$\therefore |z-1|=1$
Or,
$\sin (\theta -\alpha )=0\Rightarrow \theta -\alpha =0\Rightarrow \theta =\alpha \therefore \arg (z-1)=\alpha$