Question

For every real number $c>0,$ find all complex numbers z, satisfying the equation $z\left|z\right|+cz+i=0.$

• A. $(x,y)=(0,c\pm \frac{\sqrt{c^2-4}}{2})$
• B. $(x,y)=(0,-c\pm \frac{\sqrt{c^2-4}}{2})$
• C. $(x,y)=(0,c\pm \frac{\sqrt{c^2-3}}{2})$
• D. $(x,y)=(0,-c\pm \frac{\sqrt{c^2-3}}{2})$

Answer 1

The correct option is B $(x,y)=(0,-c\pm \frac{\sqrt{c^2-4}}{2})$

Let
$z=x+iy$
Hence ,$(x+iy)\left(\sqrt{x^2+y^2}\right)+cx+icy+i=0$
Hence the real part is
$\Rightarrow x\sqrt{x^2+y^2}+cx=0$
Hence ,$x=0$ since $|z|\ne 0$
And substituting $x=0$, we get the imaginary part as
$\Rightarrow y^2+cy+1=0$
$y=\frac{-c\pm \sqrt{c^2-4}}{2}$