If a complex number z statisfies the equation | vedclass

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Question

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}[\

Vedclass · Accepted Answer

$\sqrt 5$

Vedclass · Answer

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, $z=-2+i(-1) \Rightarrow|z|=\sqrt{5}$

If a complex number $z$ statisfies the equation $x + \sqrt 2 \,\,\left| {z + 1} \right|\,+ \,i\, = \,0,$ then $\left| z \right|$ is equal to

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