Question

The image of the complex number $\frac{2-i}{3+i}$, in the straight line $z(1+i)=\overline{z}(i-1)$, where $z$ is a complex number, is/are

• A. $\frac{-1-i}{2}$
• B. $\frac{-1+i}{2}$
• C. $\frac{i(i+1)}{2}$
• D. $\frac{-1}{1+i}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\frac{-1+i}{2}$
C $\frac{i(i+1)}{2}$
D $\frac{-1}{1+i}$

$\frac{2-i}{3+i}=\frac{(2-i)(3-i)}{9+1}=\frac{5-5i}{10}$
$=\frac{1}{2}-\frac{i}{2}$
i.e.$(\frac{1}{2},-\frac{1}{2})$ is given point.
and $z(1+i)=\overline{z}(i-1)$
Let $z=x+iy$
$\Rightarrow (z+\overline{z})+i(z-\overline{z})=0$
$\frac{z+\overline{z}}{2}+i\frac{z-\overline{z}}{2}=0$
$\Rightarrow x+i\left(iy\right)=0$
$x-y=0\Rightarrow y=x$

Reflection of $(\frac{1}{2},-\frac{1}{2})$ with respect to $y=x$ is $(-\frac{1}{2},\frac{1}{2})$
i.e. $-\frac{1}{2}+\frac{i}{2}=\frac{-1+i}{2}$
$=\frac{i^2+i}{2}=\frac{i(i+1)}{2}$
$=\frac{i(1+i{)}^2}{2(1+i)}=\frac{i(1+i^2+2i)}{2(1+i)}$
$=\frac{i^2}{1+i}=\frac{-1}{1+i}$