Complex question?

Let $z = x + iy$ with $x,y \in \mathbb{R}$ , then

$(x + iy)^2 + \sqrt{x^2 + y^2} = 0$

$x^2 - y^2 + 2xyi = - \sqrt{x^2 + y^2}$

Comparing imaginary and real parts, we find that x = y = 0 is a solution or that either x = 0 or y = 0. The case in which y = 0 gets excluded, because the left hand side would be strictly positive $(x^2)$ , while the right hand side $(-\sqrt{x^2})$ would be strictly negative. (Note that the only solution $(x= y = 0)$ is already found by other reasoning, math is extremely beautiful :-)).

We thus continue with $x= 0$ and get:

$-y^2 = - \sqrt{y^2}$

$y^4 = y^2$

$y^2 = 1$

$y = 1 \vee -1$ .

Hence we get three solutions $(0, i, -i)$ .

Let z=re^(iθ), then r^2. e^(i2θ) + r = 0, then r.e^(i2θ) = -1 = e^i(π+2 πk), so θ= π/2 + πk, where
k= 0,1,2...,and r =1.So the the solutions are z=0,i,-i.

Put z=x+iy. The given equation is then equivalent to xy=0 and x^2 + sqrt(x^2+y^2)=y^2. The only solutions are (x, y)=(0,0);(0,1);(0,-1)

@jaikirat sandhu , how is this level 4?

The complex nos. are 0, i, and -i