Question

Solve for $z$, i.e, find all complex numbers $z$ which satisfy ${\left|z\right|}^2-2iz+2c(1+i)=0$, where $c$ is real.

• A. $z=c+i(-1\pm \sqrt{1+2c-c^2})$,Where $c\in [-1,-\sqrt{2},-1+\sqrt{2}]$
• B. $z=c+i(-1\pm \sqrt{1-2c-c^2})$,Where $c\in [-1,-\sqrt{2},-1+\sqrt{2}]$
• C. $z=c+i(-1\pm \sqrt{1-2c-c^2})$,Where $c\in [1,-\sqrt{2},1+\sqrt{2}]$
• D. $z=c+i(-1\pm \sqrt{1+2c-c^2})$,Where $c\in [1,-\sqrt{2},1+\sqrt{2}]$

Answer 1

The correct option is B $z=c+i(-1\pm \sqrt{1-2c-c^2})$,Where $c\in [-1,-\sqrt{2},-1+\sqrt{2}]$

Put $z=x+iy$.
Then the given equation reduces to
$x^2+y^2-2i(x+iy)+2c(1+i)=0$
Or $(x^2+y^2+2y+2c)+2i(a-x)=0$
Equating the real and imaginary part to zero, we get
$x^2+y^2+2y+2c=0,2c-2x=0$
This gves for $x=c$ and for y we have
$c^2+y^2+2y+2c=0$ or $y^2+2y+(c^2+2c)=0$ ...(1)
Since we seek real value of y, the discriminant of the equation (1)
must be non-negative, that is
$△=4-4(c^2+2c)\ge 0\Rightarrow 1-c^2-2c\ge 0$
For the real value of c, we get
$y=\frac{-2\pm \sqrt{4(1-c^2-2c)}}{2}=-1\pm \sqrt{1-c^2-2c}$
Hence for $△\ge 0$, the original equation has two roots
$z_1=c+(-1+\sqrt{1-c^2+2c})i{z}_2=c+(-1-\sqrt{1-c^2+2c})i$
For $△=0$, we have $z_1=z_2$ that is,
there is only one solution in this case.
For $△<0$ the equation has no roots
It remain to indicate the range of $c$ over which there are solution.
So $c$ must satisfy the inequality.
$1-c^2-2c\ge 0\Rightarrow c^2+2c-1\le 0\Rightarrow {(c+1)}^2-2\le 0{(c-1)}^2\le 2\Rightarrow -\sqrt{2}\le c+1\le \sqrt{2}\Rightarrow -1-\sqrt{2}\le c\le -1+\sqrt{2}$