singular conic is two lines in $\mathbb{C}$ So given $F(z,w)=az^2+bzw+cw^2+dz+ew+f$ a singular conic I need to show if I am workin in $\mathbb{C}$ that I can write this as two line, here's what I have:
$F$ being singular means $\exists (z_0,w_0)\in V(F)$
$F(z_0,w_0)=az_0^2+bz_0w_0+cw_0^2+dz_0+ew_0+f=0$
$F_z(z_0,w_0)=2az_0+bw_0+d=0$
$F_w(z_0,w_0)=2cw_0+bz_0+e=0$
Somehow If I can deduce $d=e=f=0$ that would be great cause that guy would factor over $\mathbb{C}$, any hints?
 A: Shift $z,w$ so that $z_0=w_0=0$. Then $F(0,0)=0$ means $f=0$, $\partial_z F(0,0)=0$ means $d=0$, $\partial_w F(0,0)=0$ means $e=0$.  Can you finish from there ?
A: This is a "side answer" to the question. I would like here to explain where the condition "working in $\mathbb{C}$" plays an essential rôle.
$$F(z,w)=az^2+bzw+cw^2+dz+ew+f$$
If $a=c=0$, we have a first degree equation, and we have finished.
Let us assume that one among $a$ or $c$ is $\neq 0$. We will consider WLOG that $a \neq 0$. Dividing by $a$, we can even assume that $a=1$.
We can assume, up to a multiplication by an adequate constant that $F(z,w)$ can be written :
$$(z+v_1w+w_1)(z+v_2w+w_2)\tag{1}$$
Expanding and identifying similar coefficients, we find :
$$\begin{cases}v_1 + v_2&=&b\\v_1v_2&=&c \end{cases} \ \ \ \ \ \ \begin{cases}w_1 + w_2&=&d\\
w_1w_2&=&f\end{cases} \ \ \ \ \text{and} \ \ v_1w_2 + v_2w_1=e$$
Consider the first set of equations. You surely have seen that, when one knows the sum $S$ and the product $P$ of two numbers, there is a unique set of solutions, the roots of the quadratic equation $x^2-Sx+P=0$, and these roots can be complex (if the discriminant $S^2-4P=b^2-4c<0$). The same for the second group of equations. 
Concerning the last equation, it is automaticaly verified because, in the hypothesis, we have the factorisation (1).
