Question

For $c\ge 0,$ find number of complex numbers $z$ which satisfy the equation ${\left|z\right|}^2-2iz+2c(1+i)=0.$

Answer 1

Put $z=x+iy.$

Then equation takes the form
$x^2+y^2-2i(x+iy)+2c(1+i)=0$
$(x^2+y^2+2y+2c)+2i(c-x)=0$
Equating real and imaginary parts to zero, we get
$x^2{y}^2+2y+2c=0,2c-2x=0\therefore x=c$ ...(1)
These give $x=c$ and for $y$ on putting for $x$, we get
$y^2+2y+c^2+2c=0$
$y=\frac{-2\pm \sqrt{4(1-c^2-2c)}}{2}$ ...(2)
where $1-c^2-2c\ge 0$ for real yor $2\ge c^2+2c+1$ or ${(c+1)}^2-{\left(\sqrt{2}\right)}^2\le 0$
$\therefore -\sqrt{2}\le (c+1)\le \sqrt{2}$
$-\sqrt{2}-1\le c\le \sqrt{2}-1$
Since $c\ge 0$ we contract the above interval to. $0\le c\le \sqrt{2}-1$
I. $c=0(x,y)=(0,-1\pm 1)$ or $(0,0),(0,-2)$
II. $0<c<(\sqrt{2}-1).$
There are two solutions $x,y$ given by (1) and (2).
III. $c=\sqrt{2}-1i.e.1-c^2-2c=0,$ then from (1) and (2) there is only one solution from (1) and (2), $(x,y)=(c,-1)$
IV. For $c>\sqrt{2}-1,\Delta <0$ and hence there is no solution.