Question

If $(1+i)z=(1-i)\overline{z}$ then $z=x+iy$ is

• A. $x(1-i),xϵR$
• B. $x(1+i),xϵR$
• C. $\frac{x+1}{1+i},xϵR^{+}$
• D. None of these

Answer 1

The correct option is A $x(1-i),xϵR$

Given, $(1+i)z=(1-i)\overline{z}$
Multiply both sides by $(1-i)$
$(1+i)(1-i)z=(1-i{)}^2\overline{z}$
$\Rightarrow (1-i^2)=(1-2i+i^2)\overline{z}$
$\Rightarrow 2z=-2i\overline{z},\because i^2=-1$
$\Rightarrow z=-i\overline{z}$ ......(1)
Now let $z=x+iy\therefore \overline{z}=x-iy$
$\left(1\right)\Rightarrow x+iy=-i(x-iy)=-ix+i^{2}y=-y-ix$
Equating coefficients, $y=-x$
$\therefore z=x-ix=x(1-i),x\in R$