Complex numbers straight line proof 
Prove that the three distinct points $z_1,z_2$, and $z_3$ lie on the
  same straight line iff $z_3 - z_2 = c(z_2 - z_1)$ for some real number
  $c$ and $z$ is complex.

I know that two vectors are parallel iff one is a scalar multiple of the other, thus $z$ is parallel to $w$ iff $z = cw$. So, from that, does that mean $z_3 - z_2 = c(z_2 - z_1)$ are parallel thus making it lie on the same line?
 A: If in a complex plane $z_1$ is represented by point $A, z_2$ by $B$ and $z_3$ by $C$, then $z_3-z_2$ represents the vector $\vec {BC}$, and $z_2-z_1$ represents vector $\vec {AB}$. Thus, these points $A,B,C$ are collinear iff $\vec{AB}$ and $\vec{BC}$ are parallel as they already have one point $B$ in common $\implies \vec {BC}=c\vec{AB}\implies z_3-z_2=c(z_2-z_1)$.
Here, $c$ is taken real because when we multiply by a complex no. , it's real part gives the scaling and imaginary part rotates it. Here, we need the vectors to be parallel ,so we omit the rotation and $c$ has to be real only.
A: If $z_k=x_k+iy_k$ for $k=1,2,3$
As $z_3-z_2=c(z_2-z_1),$ 
If $c=0, z_3=z_2$ and if $z=\infty, z_2=z_1$ so $c$ non-zero finite number.
$\implies x_3-x_2+i(y_3-y_2)=c\{x_2-x_1+i(y_2-y_1)\}$
Equating the real & the imaginary parts, $$x_3-x_2=c(x_2-x_1),y_3-y_2=c(y_2-y_1)$$
So, $$\frac{y_3-y_2}{x_3-x_2}=\frac{y_2-y_1}{x_2-x_1}\implies x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$ 
Hence $z_1,z_2$ and $z_3$ are collinear.

Alternatively,
the area of the triangle with vertices $z_k=x_k+iy_k$ for $k=1,2,3$
is $$\frac12\det\begin{pmatrix}x_1&y_1 &1\\x_2 & y_2 & 1 \\ x_3 & y_3&1\end{pmatrix}$$
$$=\frac12\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ x_3-x_2 & y_3-y_2&0\end{pmatrix}$$ applying $R_1'=R_1-R_2$ and $R_3'=R_3-R_2$
$$=\frac12\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ -c(x_1-x_2) & -c(y_1-y_2)&0\end{pmatrix}$$
$$=-\frac c2\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ x_1-x_2&y_1-y_2&0\end{pmatrix}=0$$ as the 1st & the 3rd rows are identical. 
Hence $z_1,z_2$ and $z_3$ are collinear.
A: WLOG let $z_1$ be at the origin, and $z_2 -z_1 = r_1e^{i\theta_1},$ an arbitrary vector from the origin. $z_2$ is the endpoint of this vector, and we have a third point $z_3$ and want to know if the angle of $z_3-z_2 = r_2e^{i\theta_2}$ is that of $z_2-z_1.$
But $$\frac{z_3 -z_2}{z_2-z_1}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}= \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} = c. $$ Since c is real, $\theta_1 -\theta_2$ must equal zero, so they are equal.  
