Complex Number

Let $z = a + bi$ be any complex number. The above equation yields:

$(a+bi)^2 + \sqrt{a^2 + b^2} = 0;$

or $a^2 - b^2 + 2abi + \sqrt{a^2 + b^2} = 0$

which requires both the real and the imaginary parts to equal zero:

$a^2 - b^2 + \sqrt{a^2 + b^2} = 0; 2ab = 0$ .

The imaginary part is zero iff $a$ or $b = 0$ . Hence for the real part:

If $a = 0$ , then we have $-b^2 + |b| = 0 \Rightarrow b = \pm b^2.$ Taking $b = b^2$ yields $b = 0, 1$ and $b = -b^2$ yields $b = 0, -1$ , which in turn produce the ordered-pairs $(a,b) = (0,0); (0,-1); (0,1)$ .

If $b = 0$ , then we have $|a| = -a^2 \Rightarrow a = 0$ and $(a,b) = (0,0)$ is the only ordered-pair solution under this condition.

Thus, $(a,b) = (0,0); (0,-1); (0,1)$ (or $\boxed{z = 0, \pm i}$ ) are the only solutions

Use basics and always stick to the original form of the complex numbers x +yi. Thus get a quadratic or quartic equation where we face a compulsion to put x or y as 0. Hence using simpke lohic the answer comes out to be 0 i and -i. There are thus three solutions.