A problem in Beardon's Algebra and Geometry asks that you show that the equation of the line that passes through the origin and the direction $b$ is $\bar{b}z = b\bar{z}$, where $b\not=0.$
My reasoning is that since the line is in the direction of $b$ and passes through the origin, $b$ must be on the line. And so substituting both $0$ and $b$ for $z$ satisfy the equation.
My question is how to make this conform to the relation $\bar{a}z + a\bar{z} + c = 0$, derived from $|z-u|^2 = |z-v|^2$, where a line is defined by all the points equidistant from two given points $u$ and $v.$
It seems that $\bar{b} = -b$ necessarily, which is only true when $b=0.$
Please let me know what I'm missing here.
As you know, $|z - u|^2 = |z-v|^2$ expands out to $$ (v-u)\bar{z} + (\bar{v}-\bar{u})z + \frac{|u|^2-|v|^2}{2} = 0 $$ which gives $a = v-u$ and $c = (|u|^2 - |v|^2)/2$. However, $v-u$ is perpendicular to the line defined by the above equation, which means that the line is actually in the direction $b = ia$. For a line that goes through the origin, $c = 0$, and so $$ 0 = a\bar{z} + \bar{a}z +c = ib\bar{z}-i\bar{b}z \Longrightarrow b\bar{z} = \bar{b}z $$ as desired.