Using $z = x + yi$ does not reveal all solutions

Question

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?

Answer 1

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 $(1,0)$.

Answer 2

See @user10354138's answer for your question, but I just wanted to add: Another way you can do this is just using the straight-up quadratic equation on $z^2−(i+1)z+i=0$. Namely, $$z = \frac{i+1 \pm \sqrt{(1+i)^2-4i}}{2} = \frac{i+1 \pm (-1+i)}{2}$$ Therefore, $z = i$ or $z=1$, which are the two answers you got with your method.