Question

Find a complex number $z=x+iy$ which satisfies the given equation:$z+\surd 2|z+1|+i=0$

Answer 1

$z+\sqrt{2}|z+1|+i=0$

$z=x+iy$

$|z+1|=|(x+1)+iy|$

$\therefore |z+1|=\sqrt{(x+1{)}^2}+y^2$

$\therefore x+iy+\sqrt{2}(\sqrt{(x+1{)}^2}+y^2+i=0$

$\therefore (x+\sqrt{2(x+1{)}^2+2y^2)}+(iy+1)=0$

$\therefore iy+i=0;y=-1$

Also $x+\sqrt{2(x+1{)}^2+2}=0$

$\therefore x+\sqrt{2(x+1{)}^2+2}=0$

$\therefore x^2=2(x+1{)}^2+2$

$\therefore x^2=2(x^2+2x+1)+2$

$\therefore x^2=2x^2+4x+2+2$

$\therefore x^2+4x+4=0$

$\therefore (x+2{)}^2=0$

$\therefore x=-2$

$z=-2-i$

Cpmplex number is $-2-i$