Question

Consider the given expression $\left( 1+i \right)\times z=\left( 1-i \right)\bar{z}$, then $z=x+iy$ is
A. $z\left( 1-i \right),x\in R$
B. $x\left( 1-i \right),x\in R$
C. $\dfrac{x+1}{1+i},x\in {{R}^{T}}$
D. none of these

Answer 1

Hint: Rearrange the given expression and take their conjugate. Find the relation connecting z and $\bar{z}$. Put $z=x+iy$ and $\bar{z}=x-iy$ and thus get the expression for z.

Complete step-by-step answer:
We have been given the expression, $\left( 1+i \right)\times z=\left( 1-i \right)\bar{z}$.
Let us cross multiply the above expression. We get,

& \left( 1+i \right)\times z=\left( 1-i \right)\bar{z} \\\ & \Rightarrow z=\left( \dfrac{1-i}{1+i} \right)\bar{z} \\\ \end{aligned}$$ Multiply the numerator and denominator with $$\left( 1-i \right)$$. $$z=\dfrac{\left( 1-i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)}\bar{z}$$ $$\begin{aligned} & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\\ & \because {{i}^{2}}=-1 \\\ \end{aligned}$$ $$\begin{aligned} & z=\dfrac{{{\left( 1-i \right)}^{2}}}{{{1}^{2}}-{{(i)}^{2}}}\bar{z}=\left( \dfrac{1-2i+{{i}^{2}}}{1-{{i}^{2}}} \right)\bar{z} \\\ & z=\dfrac{1-2i-1}{1-(-1)}\bar{z} \\\ & z=\dfrac{-2i}{2}\bar{z}\Rightarrow z=-i\bar{z} \\\ \end{aligned}$$ Thus we got $$z=-i\bar{z}.......(1)$$ Now it’s given that $$z=x+iy$$, so $$\bar{z}=\left( x-iy \right)$$. Substitute the value of $$\bar{z}$$ in equation (1). $$\therefore \bar{z}=-i\left( x-iy \right)$$ Open the brackets and simplify the expression $$\therefore z=-ix+(-1)y$$ We know $${{i}^{2}}=-1$$. $$\begin{aligned} & =-ix-y \\\ & z=-(y+ix).......(2) \\\ & \therefore z=-i(x-iy)=-(y+ix) \\\ \end{aligned}$$ From this we can say that $$x=-y$$. $$\begin{aligned} & z=x+iy=x-ix \\\ & \therefore z=x\left( 1-i \right) \\\ \end{aligned}$$ Thus we got the values as $$z=x\left( 1-i \right),x\in R$$. Option B is the correct answer. Note: Remember that if $$z=x+iy$$, then $$\bar{z}=x-iy$$. This is some of the important portions in complex numbers. You need to remember basic geometric formulae which we have used here. Thus after getting $$z=-i\bar{z}$$, substitute the necessary values and solve it.