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4-1.Complex numbers
hard
જો સંકર સંખ્યા $z$ માટે $x + \sqrt 2 \,\,\left| {z + 1} \right|\,+ \,i\, = \,0$ હોય તો $\left| z \right|$ ની કિમત મેળવો.
A
$2$
B
$\sqrt 3$
C
$\sqrt 5$
D
$1$
(JEE MAIN-2013)
Solution
Given equation is
$z+\sqrt{2}|z+1|+i=0$
put $z=x+i y$ in the given equation.
$(x+i v)+\sqrt{2}|x+i y+1|+i=0$
$\Rightarrow x+i y+\sqrt{2}[\sqrt{(x+1)^{2}+y^{2}}]+i=0$
Now, equating real and imaginary part, we get
$x+\sqrt{2} \sqrt{(x+1)^{2}+y^{2}}=0$ and
$y+1=0 \Rightarrow y=-1$
$\Rightarrow x+\sqrt{2} \sqrt{(x+1)^{2}+(-1)^{2}}=0$
$(\because y=-1)$
$\Rightarrow \sqrt{2} \sqrt{(x+1)^{2}}+1=-x$
$\Rightarrow 2\left[(x+1)^{2}+1\right]=x^{2}$
$\Rightarrow x^{2}+4 x+4=6$
$\Rightarrow x=-2$
Thus, $z=-2+i(-1) \Rightarrow|z|=\sqrt{5}$
Standard 11
Mathematics
