Question

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

• A. $\lambda (1-i),\lambda ϵR$
• B. $\lambda (1+i),\lambda ϵR$
• C. $\frac{\lambda }{1+i^3}\lambda ϵR$
• D. $\frac{\lambda }{1-i^2}\lambda ϵR$

Answer 1

The correct option is A $\lambda (1-i),\lambda ϵR$

$(1+i)z=(1-i)\overline{z}$
Divide above equation by $(1+i)(1-i)$ we get , $\frac{z}{1-i}=\frac{\overline{z}}{1+i}$
Let, $\frac{z}{1-i}=\frac{\overline{z}}{1+i}=\lambda$ where $\lambda \in R$
$\therefore z=\lambda (1-i)$ , $\lambda \in R$
Ans: A