Question

For $c\ge 1,$ find number of complex numbers z that satisfy the equation $z+c|z+1|+i=0.$

Answer 1

$z=x+iy$
Then
$x+iy+c\sqrt{(x+1{)}^2+y^2}+i=0$
$(x+c\sqrt{(x+1{)}^2+y^2})+i(y+1)=0$
Now
Real part =0 and imaginary part=0.
Hence
$y=-1$ ...(i)
And $x+c\sqrt{(x+1{)}^2+y^2}=0$
Or
$x+c\sqrt{(x+1{)}^2+(-1{)}^2}=0$ from i.
$-x=c\sqrt{(x+1{)}^2+(-1{)}^2}$
$x^2=c^2(x+1{)}^2+c^2$
$x^2=c^2{x}^2+2c^{2}x+2c^2$
Or
$(c^2-1)x^2+2c^{2}x+2c^2=0$
$x=\frac{-2c^2\pm \sqrt{4c^4-8c^2(c^2-1)}}{2(c^2-1)}$
$x=\frac{-2c^2\pm \sqrt{8c^2-4c^4}}{2(c^2-1)}$
$x=\frac{-c^2\pm \sqrt{2c^2-c^4}}{c^2-1}$
$x=\frac{-c^2\pm c\sqrt{2-c^2}}{c^2-1}$
For real x,
$2-c^2>0$
$c^2<2$
Or
$1\le c\le \sqrt{2}$ since $c\ge 1$.
Now if c=1, x=-1, hence (-1,-1).
If $c=\sqrt{2}$, we get $x=-2$ hence $(-2,-1)$ and so on.