If (3 + i)z = (3 - i)bar{z},then the complex | vedclass

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(A) Given equation: $(3 + i)z = (3 - i)\bar{z}$.Let $z = x(3 - i)$ where $x \in R$.Then $\bar{z} = x(3 + i)$.Substituting these into the left-hand sid

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$x(3 - i), x \in R$

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(A) Given equation: $(3 + i)z = (3 - i)\bar{z}$.Let $z = x(3 - i)$ where $x \in R$.Then $\bar{z} = x(3 + i)$.Substituting these into the left-hand side $(LHS)$: $(3 + i)z = (3 + i)x(3 - i) = x(3^2 + 1^2) = 10x$.Substituting these into the right-hand side $(RHS)$: $(3 - i)\bar{z} = (3 - i)x(3 + i) = x(3^2 + 1^2) = 10x$.Since $LHS = RHS$,the complex number $z$ is of the form $x(3 - i)$ for $x \in R$.

If $(3 + i)z = (3 - i)\bar{z}$,then the complex number $z$ is

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