Question

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

• A. $t(1-i),tϵR$
• B. $t(1+i),tϵR$
• C. $\frac{t}{1+i},tϵR$
• D. none of these

Answer 1

The correct option is A $t(1-i),tϵR$

Let $z=x+iy$

$\therefore \overline{z}=x=-iy$

$(1+i)z=(1+i)(x+iy)=x+iy+ix-y$

$(1-i)\overline{z}=(1-i)(x-iy)=x-iy-ix-y$

$(1+i)z=(1-i)\overline{z}$

$(1+i)z=(1-i)\overline{z}$

$⟹x+iy+ix-y=x-iy-ix-y$

$2i(x+y)=0$

$x+y=0$

$x=-y$

$\therefore z=x-xi=x(1-i)$

$z=t(1-i),t\in R$