Question

If $z+\sqrt{2}|z+1|+i=0$ and $z=x+iy$, then

• A. $x=-2$
• B. $x=2$
• C. $y=-2$
• D. $y=1$

Answer 1

Answer: (A)

$x=-2$

On putting $z=x+iy$, We get
=> $x+iy+\sqrt{2(x+1{)}^2+2y^2}+i=0$
=> $x+\sqrt{2(x+1{)}^2+2y^2}+i(y+1)=0$
=> On comparing real part and imaginary part, We get
=> $x+\sqrt{2(x+1{)}^2+2y^2})=0$ and $y+1=0$
So, $y=-1$
and , $\sqrt{2(x+1{)}^2+2y^2})=-x$
=> $2(x+1{)}^2+2=x^2$
=> $x=-2$