Show that vectors defined by complex numbers $z_1$ and $z_2$ are parallel if and only if $\operatorname{Im}(z_1 \bar{z_2})=0$.
Does the proofing use the fact that in $\mathbb{R}^2$ the cross product of two vectors is scalar as in $\mathbf{u}\times\mathbf{v}=u_1 v_2 -u_2 v_1=0 \to \text{parallel}$.
Also $\operatorname{Im}(z_1 \bar{z_2})=0$ if $z_1=z_2$ so then the expression reduces to $\operatorname{Im}(x^2 +y^2+\underbrace{i(-xy+xy)}_{=0})=0$.
So two vectors in $\mathbb{R}^2$ are parallel if $u_1 v_2 -u_2 v_1 = 0 $ then for the complex numbers it has to be that $u_1=y_1=x$ and $u_2=v_2=y$?
I'm not sure if this is the way to show the proof though.
Best Answer
I'd rather go this way: write $\;z_1=(a,b)\;,\;\;z_2=(x,y)\;,\;\;a,b,x,y\in\Bbb R\;$ .
Then, $\;z_1\,,\;z_2\;$ are parallel iff
$$\exists\;k\in\Bbb R\;\;s.t.\;\;(x,y)=k(a,b)=(ka,kb)\iff$$
$$ z_1\overline z_2=(a,b)(ka,-kb):=\left(k(a^2+b^2)\;,\;k(-ab+ab)\right)=\left(k(a^2+b^2)\;,\;0\right)\iff$$
$$\iff\text{Im}\,(z_1\overline z_2)=0$$
Note: the last backward implication ($\Leftarrow$) is not obvious but can be deduced as follows:
$$\text{Im}\,(z_1\overline z_2)=0 \implies \text{Im}\,((ax+by,bx-ay))=0\implies bx-ay=0\implies \frac{x}{a}=\frac{y}{b}=k$$