How to solve that quadratic complex equation? z2−1z2+z+1 So, there cant

Step 1Writing$z=x+iy$then$z^2+z+1=(x+iy{)}^2+(x+iy)+1$$=(x^2-y^2+x+1i(2xy+y)$Step 2If$(x^2-y^2+x+1)+i(2xy+y)=0$with x,y real numbers, then you need$x^2-y^2+x+1=0$and$2xy+y=0$You solve these the way you usually solve equations for real numbers.So, for example, you have$0=2xy+y=(2x+1)y$So either $y=0$ or $2x+1=0$. If $y=0$, then the first equation reduces to $x^2+x+1=0$, which has no real solutions, so there are no solutions with $y=0$. If $y\ne 0$, then $2x+1=0$, so $x=-\frac{1}{2}$. Plugging into the first equation, we get$0=\frac{1}{4}-y^2-\frac{1}{2}+1=-y^2+\frac{3}{4}$so you get that$y^2=\frac{3}{4}$or$y=\pm \frac{\sqrt{3}}{2}$So the two solutions are$z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$and$z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$Step 3The acrobatics from step 2 are unnecessary. You don't have to decompose into real and imaginary parts, because the quadratic formula works for complex numbers! (Provided you take complex square roots). Since$z^2+z+1=(z+\frac{1}{2}{)}^2+\frac{3}{4}$(by completing the square), then this is zero if and only if$z+\frac{1}{2}=\sqrt{-\frac{3}{4}}$if and only if$z=-\frac{1}{2}+\frac{\sqrt{-3}}{2}$You may recognize this as exactly what you get from the quadratic formula applied to$z^2+z+1$and you may also recognize them as the solutions you get if you go through the contorsions of step 2 above. So just find the two complex square roots of -3, and rejoice! (The quadratic formula works even if the coefficients of the quadratic are complex numbers, instead of real numbers).

The maximal domain of definition is,as you imply, the complement in $\mathbb{C}$ of the set of roots of the polynomial $z^2+z+1$ Therefore to find the bad points, we need to solve the equation $z^2+z+1=0$ I am pretty sure, now, that you know how to solve quadratic equations, no?