Question

The solutions of the equation $z^2+|z|=0$ are

• A. $z=0$
• B. $z=i$
• C. $z=-i$
• D. $z=1+i$

Answer 1

Answer: (A), (B), (C)

$z=-i$

Let $z=x+iy$
$\therefore |z|=\sqrt{x^2+y^2}$

$z^2+|z|=(x+iy{)}^2+\sqrt{x^2+y^2}=0$
$\Rightarrow x^2-y^2+2xyi+\sqrt{x^2+y^2}=0+0i$
Equating real and imaginary parts from both sides
$x^2-y^2+\sqrt{x^2+y^2}=0...\left(1\right)$

and $2xy=0$
$\Rightarrow x=0\text{or}y=0$

If $y=0$
Eqn. $\left(1\right)\to x^2+\sqrt{x^2}=0$
$\Rightarrow x^2+|x|=0$
When $x\ge 0$
$x^2+x=0\Rightarrow x(x+1)=0$
$\Rightarrow x=0\text{or}x=-1\left(\text{not possible}\right)$
$\therefore y=0\Rightarrow x=0\Rightarrow z=0$

When $x<0$
$x^2-x=0\Rightarrow x(x-1)=0$
$\Rightarrow x=0\text{or}x=1\left(\text{not possible}\right)$

If $x=0$
Eqn. $\left(1\right)\to -y^2+|y|=0$
When $y\ge 0$
$-y^2+y=0\Rightarrow y(y-1)=0$
$\Rightarrow y=0\text{or}y=1$
$\therefore x=0\Rightarrow y=1\Rightarrow z=i$

When $y<0$
$-y^2-y=0\Rightarrow y(y+1)=0$
$\Rightarrow y=0\text{or}y=-1$
$\therefore x=0\Rightarrow y=-1\Rightarrow z=-i$