<s>[INST] 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.
 [/INST] 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.
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