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Q.
If a complex number $z$ statisfies the equation $z+\sqrt{2}\left|z+1\right|+i=0,$ then $\left|z\right|$ is equal to:
JEE MainJEE Main 2013Complex Numbers and Quadratic Equations
Solution:
Given equation is
$z+\sqrt{2}\left|z+1\right|+i=0$
put $z = x + iy$ in the given equation.
$\left(x+iy\right)+\sqrt{2}\left|x+iy+1\right|+i=0$
$\Rightarrow x+iy+\sqrt{2}\left[\sqrt{\left(x+1\right)^{2}+y^{2}}\right]+i=0$
Now, equating real and imaginary part, we get
$x+\sqrt{2}\sqrt{\left(x+1\right)^{2}+y^{2}}=0$ and
$y+1=0 \Rightarrow y=-1$
$\Rightarrow x+\sqrt{2}\sqrt{\left(x+1\right)^{2}+\left(-1\right)^{2}}=0\,\left(\because y=-1\right)$
$\Rightarrow \sqrt{2}\sqrt{\left(x+1\right)^{2}+1}=-x$
$\Rightarrow 2\left[\left(x+1\right)^{2}+1\right]=x^{2}$
$\Rightarrow x^{2}+4x+4=0$
$\Rightarrow x=-2$
Thus, $z=-2+i\left(-1\right) \Rightarrow \left|z\right|=\sqrt{5}$