Find all complex numbers satisfying the equation $\bar{z}+1=iz^2+|z|^2$ Find all complex numbers z satisfying $$\bar{z}+1=iz^2+|z|^2$$
where $i=\sqrt{-1}$
I only know one way i.e. assuming $z=x+iy$ but that process is very cumbersome. I don't know how to proceed otherwise with a shorter approach.
 A: I do not find it so cumbersome to replace $z=x+iy$ if you write out the details (it looks like it might be a mess, and, without trying it out, it's hard to see where the problem simplifies).  Substituting $z=x+iy$ into the equation gives
$$
(x+1)+(-y)i=(x^2+y^2-2xy)+(x^2-y^2)i=(x-y)^2+(x^2-y^2)i.
$$
Therefore
\begin{align*}
x+1&=(x-y)^2\\
-y&=x^2-y^2
\end{align*}
From here, you have that $x^2=y^2-y$.  You could, at this point, take the first equation and rewrite it as
$$
x+2xy=x^2+y^2-1.
$$
Factoring the LHS gives $x(1+2y)$ and substituting the formula for $x^2$ into the RHS gives $2y^2-y-1=(2y+1)(y-1)$.  So, this equation simplifies to:
$$
x(1+2y)=(1+2y)(y-1).
$$
Moving all the terms to one side, we get
$$
(1+2y)(y-x-1)=0.
$$
Therefore, either $1+2y=0$ or $y=1+x$.  In the first case, $y=-\frac{1}{2}$, we can find $x$ via the equation $x^2=y^2-y$, and then check our answers are right in the original expression.  If $y=1+x$, you can, again, substitute into $x^2=y^2-y$ to get a formula for $x$ (and then $y$).  Whichever of these solutions satisfies the original equation are the answers that you're looking for.
A: setting $$z=x+iy$$ then we get (after some algebra)
$$x+1-iy=i(x^2-y^2)+(x-y)^2$$
can you proceed?
A: The answer is$$z=i,i\omega,i\omega^2$$where $\omega,\omega^2$ are the two complex cube roots of unity.
Putting $z=x+iy$,we get
$$(x-y)^2=1+x$$and $$x^2-y^2=-y$$
Let $A=x+y$ and also let $B=x-y$
Substituting in $$(x-y)^2=1+x$$ we get $$B^2=\frac{A+B}{2}$$i.e.$$A=2B^2-B-2$$
Substituting in $$x^2-y^2=-y$$ we get $$AB=\frac{B-A}{2}$$ i.e.$$2AB=B-A$$ Thus $$2(2B^2-B-2)B=b-2B^2+B+2$$i.e $$2B^3-3B-1=0$$ Thus $$(B+1)(2B^2-2B-1)=0$$
Case I$$B=-1$$
Now $$A=2B^2-B-2=1$$
$$x=\frac{A+B}{2}=0;y=\frac{A-B}{2}=1$$Thus
$$z=i$$
Case II$$2B^2-B-1=0$$Now $$A=2B^2-B-2$$Thus $$A+B+2-2B-1=0$$i.e$$A-B=-1$$Thus $$y=-\frac{1}{2}$$
Also $$x^2-y^2=-y$$and therefore $$x^2=\frac{3}{4}$$
$$z=\pm\frac{\sqrt{3}}{2}-\frac{i}{2}=i\omega,i\omega^2$$
Finally one has three solutions $$z=i,i\omega,i\omega^2$$
where $\omega,\omega^2$ are the two complex cube roots of unity.
