For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$ 
Find the roots of:
$z^2 - 3z + (3 - i) = 0$

$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?

$$9y^2+6y+1-4y^4-18y^2-6y+12y^2=0$$
$$-4y^2+3y^2+1=0`×(-1)$$
$$4y^2-3y^2-1=0$$
$$(4y^2+1)(y^2-1)=0$$
$$y^2=1$$
$$y=+1, x=(3+1)/2= 2$$
$$y=-1, x=(-3+1)/-2=1$$
$$z=x+yi$$
$$z1=2+i$$
$$z2=1-i$$
