Consider the following geometry problem:
Find $a\in\mathbb{C}$ so that the equation of the line through $−2 + i$ and $−2i$ is $$z\bar{a} − a\bar{z} = 8i$$ where the bar represents the complex conjugate.
I feel like I am almost there, but seem to be missing something. Here is what I have so far.
Attempt:
Let $a$ be the point on the line $[b,c]$ such that $z\bar{a} − a\bar{z} = 8i$. Then \begin{align} b=-2i&\Rightarrow \bar{b}=2i\\ c=-2+i&\Rightarrow \bar{c}=-2-i \end{align} Now, we are given 2 points and must find the third point, denoted $a$. I believe this would be the same as showing linear dependence. That is show that \begin{equation} det\left|\begin{array}{l} a&\bar{a}&1\\ b&\bar{b}&1\\ c&\bar{c}&1\end{array}\right|=0 \end{equation} With a bit of linear algebra we find the determinant gives us $$(b-a)(\bar{c}-\bar{a})-(\bar{b}-\bar{a})(c-a)=0$$ and making our substitutions for $b,\bar{b},c$ and $\bar{c}$ we have the following expression
$$(-2i-a)(-2-i-\bar{a})-(2i-\bar{a})(-2+1-a)=0.$$
Expanding this we get
$$-4-2a+2\bar{a}-6ai=8i$$
and with some rearranging we have
$$4-2\bar{a}-(2-6i)a=8i.$$
Now, this does look similar to what I am trying to find, but I'm not sure I see the $z$ and $\bar{z}$. I am wondering if there is a step at the beginning I missed, or is there something I need to do next to get the result I require.
Let $z=x+iy, a=c-id$ where $x,y,c,d$ are real
$8i=z\bar a-\bar za=2i(dx+cy)\iff dx+cy=4$
If $z=-2i;x=0,y=-2\implies -2c=4\iff c=?$
If $z=-2+i;x=-2,y=1\implies-2d+c=4$
Can you take it from here?