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?

$$\begin{cases}
 x^2-y^2-3x+3=0 \\
 2xy-3y-1=0 
 \end{cases}$$
$$x = \frac {3y+1}{2y} \Rightarrow \left(\frac {3y+1}{2y}\right)^2-y^2-3 \cdot \frac {3y+1}{2y}+3=0 \Rightarrow$$
$$\Rightarrow (3y+1)^2-4y^4-3(3y+1)\cdot 2y+12y^2=0 \Rightarrow$$
$$\Rightarrow 4y^4-3y^2-1=0$$
Substitute $~y^2=t~$ and solve for $t$ over Reals .
