A) \[x\,(3-i),\,x\in R\]
B) \[\frac{x}{3+i},\,x\in R\]
C) \[x(3+i),\,x\in R\]
D) \[x(-3+i),\,x\in R\]
Correct Answer: A
Solution :
Given: \[(3+i)z=(3-i)\bar{z}\] Let \[z=x(3-i)\], \[x\in R\] L.H.S. = \[(3+i)z\] = \[(3+i)\,x\,(3-i)\] = \[x\,(3+i)\,(3-i)\,=x\,[{{(3)}^{2}}+{{1}^{2}}]=10x\] R.H.S. = \[(3-i)\bar{z}=(3-i)\,x\,(3+i)=x\,[{{3}^{2}}+{{1}^{2}}]=10x\] Hence, L.H.S. = R.H.S. \[\because \] \[z=x(3-i)\] satisfies the equation, then \[z=x(3-i)\], where x is a real number.
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