Comparison of representation of complex numbers - point vs. vector

Question

Please vet for errors, & approach. Please edit title, if needed.

Source: book by I-Hsiung Lin, pg. #7, problem #4.

Let $z_1, z_2, z_3$ be three complex numbers, non-collinear when considered as points. Fix any $z_0$ & let $w_1 = z_0z_1, w_2 = z_0z_2, w_3 = z_0z_3$.
(a) Show that $w_1, w_2, w_3$ are non-collinear.
(b) Graph triangles $\bigtriangleup z_1z_2z_3$ & $\bigtriangleup w_1w_2w_3$. Compare their corresponding angles & areas.

(a) As given on mse, here & copied (with modifications needed for $3$ vectors instead of two) too, as below :
If in a complex plane $z_1$ is represented by point $A, z_2$ by $B$ and $z_3$ by $C$, & let chose a point $O$, as $z_0$ then $w_3=z_3-z_0$ represents the vector $\vec {OC}$, $w_2=z_2-z_0$ represents vector $\vec {OB}$, & $w_1=z_1-z_0$ represents vector $\vec {OA}$. Thus, these points $A,B,C$ are collinear iff $\vec{OA}$, $\vec{OB}$ and $\vec{OC}$ are parallel if:
(i) the slopes of vectors are the same, & have a common point too.
But, to prove no common point exists among the $3$ vectors is difficult for me, & request guidance.
Might be the book wants an actual example. So, let the $3$ non-collinear points in complex domain be:
Let $z_1 = 1+3i, z_2 = 2 +4i, z_3 = 3+5i$. But, then one would before-hand check for the non-collinearity by either checking for slopes, or for non-zero area as in (ii) below.

(ii) easier option is to prove that the area of the triangle formed by $3$ vectors $w_1, w_2, w_3$ is not zero.
Here, have another lack, can only take area of triangle with vertices, rather than vectors. Else, need at least one angle in the triangle. Need compute angle by the 'law of cosines' as stated here, as all $3$ side lengths can be computed. Let the side length of vector $w_1 = a, w_2 =b, w_3 = c$. By finding any one angle, say $\alpha$ (let it be opposite to $w_3$ with length $c$), by having $c^2 = a^2+b^2-2ab\cos\alpha$ can then compute area of triangle using vectors by taking $A = 1/2\cdot a\cdot b\cdot \sin\alpha$.
But, the above approach will not be different than using $3$ points instead, & finding area using determinant; as the origin of all $3$ vectors $w_1, w_2, w_3$ is the same/common.
However, the main issue is that am unable to prove non-collinearity of the $3$ vectors by (ii). The reason is: it is possible that the area is zero , but still no common point exists between the three vectors. So, only (i) approach seems to prove the same.

(b) Graphing is meaningless without actual coordinates, so leave that part.

For $\bigtriangleup z_1z_2z_3$ with $3$ vertices $z_1, z_2, z_3$,
the area of the triangle with vertices $z_k=x_k+iy_k$ for $k=1,2,3$ $$\frac12\det\begin{pmatrix}x_1&y_1 &1\\x_2 & y_2 & 1 \\ x_3 & y_3&1\end{pmatrix}$$ Also, as the $3$ points are given to be non-collinear, the determinant is non-zero.
To find the angles of the sides of $\bigtriangleup z_1z_2z_3$ can use the law of cosines as in (a) (ii) above.

Regarding finding the angles and areas of $\bigtriangleup w_1w_2w_3$, can follow approach (a) (ii) above.

Answer 1

This problem becomes simpler if you think geometrically rather than algebraically.

In geometric terms, multiplication by a complex number $z_0=re^{i\theta}$ is an expansion/contraction by a factor $r$ followed by a rotation through an angle $\theta$. Both of these transformations are linear, so they preserve straight lines in the complex plane. So if $z_1, z_2, z_3$ are not collinear then their images $w_1, w_2, w_3$ after multiplication by $z_0$ will not be collinear either (as long as $z_0 \ne 0$).

Similarly, the sides of the triangle are $z_1-z_2, z_2-z_3, z_3-z_1$ and these are also expanded/contracted by a factor $r$ and rotated through an angle $\theta$. The rotation does not change the area of the triangle, but the expansion/contraction does change the area - by what factor ?