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.

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.
