<s>[INST] Using $z = x + yi$ does not reveal all solutions The question is to find all solutions in $\mathbb{C}$ of the equation: $$z^2 - (i+1)z + i = 0$$
After expanding with $z = x + yi$, I get $(x^2 - y^2+y-x) + (2xy -x-y+1)i = 0$.
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.
Now, we plug that into $(2xy -x-y+1) = 0$ .
Since $x,y \in \mathbb{R}$,  this only gives the solution $x=  0, y = 1$.
However, there is another solution which is $z=1$. Why this solution does not appear with this method?
 [/INST] 
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.

That is your mistake here.  $0=x^2-y^2+y-x=(x-y)(x+y-1)$ so $x=y$ or $x=1-y$, not $y-1$.
Then substituting into the second equation $2xy-x-y+1=0$ gives $(x,y)=(0,1)$ or $(1,0)$.
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