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If a complex number $z$ statisfies the equation $x+\sqrt{2}|z+1|+i=0$, then $|z|$ is equal to :
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1767 Upvotes
Verified Answer
The correct answer is:
$\sqrt{5}$
$\sqrt{5}$
Given equation is
$$
\begin{aligned}
& z+\sqrt{2}|z+1|+i=0 \\
& \text { put } z=x+i y \text { in the given equation. } \\
& (x+i y)+\sqrt{2}|x+i y+1|+i=0 \\
& \Rightarrow x+i y+\sqrt{2}\left[\sqrt{(x+1)^2+y^2}\right]+i=0
\end{aligned}
$$
Now, equating real and imaginary part, we get
$$
\begin{aligned}
& x+\sqrt{2} \sqrt{(x+1)^2+y^2}=0 \text { and } \\
& y+1=0 \Rightarrow y=-1 \\
& \Rightarrow x+\sqrt{2} \sqrt{(x+1)^2+(-1)^2}=0 \\
& \Rightarrow \sqrt{2} \sqrt{(x+1)^2+1}=-x \\
& \Rightarrow 2\left[(x+1)^2+1\right]=x^2 \\
& \Rightarrow x^2+4 x+4=0 \\
& \Rightarrow x=-2
\end{aligned}
$$
Thus, $z=-2+i(-1) \Rightarrow|z|=\sqrt{5}$
$$
\begin{aligned}
& z+\sqrt{2}|z+1|+i=0 \\
& \text { put } z=x+i y \text { in the given equation. } \\
& (x+i y)+\sqrt{2}|x+i y+1|+i=0 \\
& \Rightarrow x+i y+\sqrt{2}\left[\sqrt{(x+1)^2+y^2}\right]+i=0
\end{aligned}
$$
Now, equating real and imaginary part, we get
$$
\begin{aligned}
& x+\sqrt{2} \sqrt{(x+1)^2+y^2}=0 \text { and } \\
& y+1=0 \Rightarrow y=-1 \\
& \Rightarrow x+\sqrt{2} \sqrt{(x+1)^2+(-1)^2}=0 \\
& \Rightarrow \sqrt{2} \sqrt{(x+1)^2+1}=-x \\
& \Rightarrow 2\left[(x+1)^2+1\right]=x^2 \\
& \Rightarrow x^2+4 x+4=0 \\
& \Rightarrow x=-2
\end{aligned}
$$
Thus, $z=-2+i(-1) \Rightarrow|z|=\sqrt{5}$
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