Question

For every real number c $\ge$ 0, the complex numbers z which satisfy the equation $|z{|}^2-2iz+2c(1+i)=0$ is $c+i(-1\pm \sqrt{1-c^2-2c})$ for $0\le c\le \sqrt{2}-1$ and for $c>\sqrt{2}-1$ there is no solution. If this is true enter 1, else enter 0.

• A. 1

Answer 1

The correct option is A 1
Let $z=x+iy$
$\therefore |z{|}^2=x^2+y^2$
$\therefore x^2+y^2-2i(x+iy)+2c(1+i)=0$
$(x^2+y^2+2y+2c)+i(-2x+2c)=0$
$(x^2+y^2+2y+2c)+i(-2x+2c)=0$
Comparing the real and imaginary parts, we get
$x^2+y^2+2y+2c=0$ ......... (1)
and $-2x+2c=0$ .......... (2)
from (1) & (2)
$y^2+2y+c^2+2c=0$
$\Rightarrow y=\frac{-2\pm \sqrt{4-4(c^2+2c)}}{2}$
$=-1\pm \sqrt{(1-c^2-2c)}$
$\because$ x and y are real,
$\therefore 1-c^2-2c\ge 0$
or $c^2+2c+1\le 2$
$(c+1{)}^2\le (\sqrt{2}{)}^2$
$\therefore -\sqrt{2}-1\le c\le \sqrt{2}-1$
$\therefore 0\le c\le \sqrt{2}-1$ ($\because$ given $c\ge 0$)
$\therefore$ Hence the solution is
$z=x+iy=c+i(-1\pm \sqrt{1-c^2-2c})$ for $0\le c\le \sqrt{2}-1$
and $z=x+iy\equiv$ no solution for $c>\sqrt{2}-1$
Ans: 1