A) 2
B) \[\sqrt{3}\]
C) \[\sqrt{5}\]
D) \[1\]
Correct Answer: C
Solution :
Given equation is \[z+\sqrt{2}|z+1|+i=0\] put \[z=x+iy\] in the given equation. \[(x+iy)+\sqrt{2}|x+iy+1|+\,i=0\] \[\Rightarrow \] \[x+iy+\sqrt{2}\left[ \sqrt{{{(x+1)}^{2}}+{{y}^{2}}} \right]+i=0\] Now, equating real and imaginary part, we get \[x+\sqrt{2}\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=0\]and \[y+1=0\Rightarrow y=-1\] \[\Rightarrow \] \[x+\sqrt{2}\sqrt{{{(x+1)}^{2}}+{{(-1)}^{2}}}=0\] \[(\because \,y=-1)\] \[\Rightarrow \] \[\sqrt{2}\sqrt{{{(x+1)}^{2}}+1}=-x\] \[\Rightarrow \] \[2[{{(x+1)}^{2}}+1]={{x}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+4x+4=0\] \[\Rightarrow \] \[x=-2\] Thus, \[z=-2+i(-1)\Rightarrow |z|\,=\sqrt{5}\]
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