Question

If the imaginary part of the complex number
$(z-1)(\cos \alpha -i\sin \alpha )+(z-1{)}^{-1}(\cos \alpha +i\sin \alpha )$
is zero, then

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

Answer 1

Answer: (B), (D)

The correct options are
B $|z-1|=1$
D $arg(z-1)=\alpha$

$(z-1)e^{-i\alpha }=\frac{e^{i\alpha }}{z-1}$
$(z-1{)}^2=e^{2i\alpha }$
$(z-1)=e^{i\alpha }$
Hence argument of $(z-1)=\alpha$
Now
$\overline{z-1}=\frac{1}{z-1}$
Therefore multiplying both sides with $\overline{z-1}$ we get
$|z-1{|}^2=|e^{2i\alpha }|$
$|z-1{|}^2=1$
$|z-1|=1$
Hence, option 'D' is correct.