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4-1.Complex numbers
easy
જો $(3 + i)z = (3 - i)\bar z,$તો સંકર સંખ્યા $z$ મેળવો.
A
$x\,(3 - i),\,x \in R$
B
$\frac{x}{{3 + i}},\,x \in R$
C
$x(3 + i),\,x \in R$
D
$x( - 3 + i),\,x \in R$
Solution
(a) 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, $L.H.S. = R.H.S.$
$z = x(3 – i)$ satisfies the equation, then $z = x(3 – i)$, where $x$ is a real number.
Standard 11
Mathematics
