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,
\[\begin{align}
  & \left( 1+i \right)\times z=\left( 1-i \right)\bar{z} \\
 & \Rightarrow z=\left( \dfrac{1-i}{1+i} \right)\bar{z} \\
\end{align}\]
Multiply the numerator and denominator with \[\left( 1-i \right)\].
\[z=\dfrac{\left( 1-i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)}\bar{z}\] \[\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\
 & \because {{i}^{2}}=-1 \\
\end{align}\]
\[\begin{align}
  & z=\dfrac{{{\left( 1-i \right)}^{2}}}{{{1}^{2}}-{{(i)}^{2}}}\bar{z}=\left( \dfrac{1-2i+{{i}^{2}}}{1-{{i}^{2}}} \right)\bar{z} \\
 & z=\dfrac{1-2i-1}{1-(-1)}\bar{z} \\
 & z=\dfrac{-2i}{2}\bar{z}\Rightarrow z=-i\bar{z} \\
\end{align}\]
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 \[{{i}^{2}}=-1\].
\[\begin{align}
  & =-ix-y \\
 & z=-(y+ix).......(2) \\
 & \therefore z=-i(x-iy)=-(y+ix) \\
\end{align}\]
From this we can say that \[x=-y\].
\[\begin{align}
  & z=x+iy=x-ix \\
 & \therefore z=x\left( 1-i \right) \\
\end{align}\]
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.