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?
 A: Use $(z-a)(z-b)=z^2-(a+b)+ab=0$ to get 
$$
(a+b)=3 \tag{1}
$$ 
$$ 
ab=(3-i) \tag{2}
$$ 
From (1) you get $\Im(a)=-\Im(b)$. So $(a_r+ia_i)(b_r-ia_i)=(a_rb_r+a_i^2)+i(-a_r+b_r)a_i=(3-i)$, thus $a_rb_r+a_i^2=3$ and $(b_r-a_r)a_i=-1$. Now you can try a few values like ...
$a_i=1$ and figure out that $a_r=2$ and $b_r=1$.
So you finally rewrite it as $(z-(2+i))(z-(1-i))$.

You could also just use the standard way (as proposed by anon) to get:
$$
\frac{3\pm\sqrt{9-4(3-i)}}{2}=\frac{3\pm\sqrt{-3+4i}}{2}.
$$
Now note that $(i(2-i))^2=-3+4i$, so
$$ \frac{3\pm i(2-i)}{2}=\frac{3\pm (2i+1)}{2}=\frac{3\pm 1 }{2} \pm i $$
A: $$\begin{cases}
 x^2-y^2-3x+3=0 \\
 2xy-3y-1=0 
 \end{cases}$$
$$x = \frac {3y+1}{2y} \Rightarrow \left(\frac {3y+1}{2y}\right)^2-y^2-3 \cdot \frac {3y+1}{2y}+3=0 \Rightarrow$$
$$\Rightarrow (3y+1)^2-4y^4-3(3y+1)\cdot 2y+12y^2=0 \Rightarrow$$
$$\Rightarrow 4y^4-3y^2-1=0$$
Substitute $~y^2=t~$ and solve for $t$ over Reals .
A: $$9y^2+6y+1-4y^4-18y^2-6y+12y^2=0$$
$$-4y^2+3y^2+1=0`×(-1)$$
$$4y^2-3y^2-1=0$$
$$(4y^2+1)(y^2-1)=0$$
$$y^2=1$$
$$y=+1, x=(3+1)/2= 2$$
$$y=-1, x=(-3+1)/-2=1$$
$$z=x+yi$$
$$z1=2+i$$
$$z2=1-i$$
