Question

Let $z$ is a complex number and $\overline{z}$ is conjugate of $z$. If $(1+i)z=(1-i)\overline{z},$ then the value of $z+i\overline{z}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $-i$

Answer 1

The correct option is C $0$

Given : $(1+i)z=(1-i)\overline{z}$
Now, let $z=a+ib$, so
$z+iz=\overline{z}-i\overline{z}\Rightarrow z-\overline{z}+i(z+\overline{z})=0\Rightarrow 2ib+2ia=0\Rightarrow b=-a$
Therefore,
$z=a-ia=a(1-i)\overline{z}=a(1+i)\Rightarrow z+i\overline{z}=a[1-i+i+i^2]\therefore z+i\overline{z}=0(\because i^2=-1)$