Reflection of a line $\bar az + a\bar z = 0$ in real axis?

Question

Why is reflection of a line $\bar az + a\bar z = 0$ given by $\bar a \bar z + az = 0$? I know that for a complex number k its reflection in real axis is $\bar k$ but in this case how can I apply that fact? In this case does $z$ become $\bar z$ and $\bar z$ become $z$? If so why?

Answer 1

Let $z=x+yi$, the reflection about the real axis is $x-yi$.

That is, on the Argand diagram, $(x,y)$ will be mapped to $(x,-y)$.

Fix $z=(x,y)$ to be a point on $\bar{a}z+a\bar{z}=0$, its reflection is $\bar{z}=(x,-y)=(\hat{x}, \hat{y})$. Let's consider what are the rules that $\hat{z}=(\hat{x}, \hat{y})$ has to satisfy. We know that $\bar{\hat{z}}$ has to lie on the line $\bar{a}z+a\bar{z}=0$. Hence, $\bar{a}\bar{\hat{z}}+a\bar{\bar{\hat{z}}}=0$, that is we have

$$\bar{a}\bar{\hat{z}}+a\hat{z}=0$$

Hence the reflection lies on the line

$$\bar{a}\bar{z}+az=0.$$

Note that using the same argument, the reflection of $f(z)=0$ is $f(\bar{z})=0$.

Answer 2

Because the map $$z\mapsto \overline{z}$$ is reflection across x-line.