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4-1.Complex numbers
hard
જો સંકર સંખ્યા $z$ ની કિમત $1 + i\alpha$, $\alpha \in R$ તથા $z^2\, = x + iy$ હોય તો
A
$y^2 -4x +2\,=0$
B
$y^2 +4x -4\,=0$
C
$y^2-4x -4\,=0$
D
$y^2 +4x +2\,=0$
(JEE MAIN-2014)
Solution
Let $z=1+i \alpha, \alpha \in R$
$z^{2}=(1+i \alpha)(1+i \alpha)$
$x+i y=\left(1+2 i \alpha-\alpha^{2}\right)$
On comparing real and imaginary parts, we get
$x=1-\alpha^{2}, y=2 \alpha$
Now, consider option $(b)$, which is $y^{2}+4 x-4=0$
$LHS$ : $y^{2}+4 x-4$
$=(2 \alpha)^{2}+4\left(1-\alpha^{2}\right)-4$
$=4 \alpha^{2}+4-4 \alpha^{2}-4$
$=0=\mathrm{R} . \mathrm{HS}$
Hence, $y^{2}+4 x-4=0$
Standard 11
Mathematics
