JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If a complex number z satisfies the equation \[z+\sqrt{2}\left| z+1 \right|+i=0,\operatorname{then}\left| z \right|\] is equal to :     JEE Main  Online Paper (Held On 22 April 2013)

    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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