Let $\bar{b}z+b\bar{z}=c,b\neq0$ be a line in the complex plane. If a point $z_1$ is the reflection of a point $z_2$ through the line, then prove that $\bar{z}_1b+z_2\bar{b}=c$
My Attempt $$ \bar{b}z+b\bar{z}-c=0\\ d_1=d_2\implies\frac{\bar{b}z_1+b\bar{z}_1-c}{2|b|}=\frac{\bar{b}z_2+b\bar{z}_2-c}{2|b|}\\ \bar{b}z_1+b\bar{z}_1=\bar{b}z_2+b\bar{z}_2\\ $$
How do I proceed further and prove the required expression ?
Since $z_1=z_2+db$ for some real $d$, the combination $\bar z_1b+z_2\bar b=\bar z_2b+db\bar b+z_2\bar b$ which is real, because it is the sum of two conjugate numbers $\bar z_2b+z_2\bar b$ and the real number $db\bar b=d|b|^2$. Therefore you can remove the real part $\mathcal{R}$ from your last relation. You need to use both conditions $(a)$ and $(b)$ I mentioned in my previous comment, since both together are equivalent to your condition of being reflected images w.r.t. the given line.