In a book I am reading, symmetry about a curve in complex plane is defined as follows:
Let $F(x,y)=0$ be a simple curve. Then points $z, z_0$ are symmetric about this curve iff
$ F \left( \dfrac{z+\bar{z_0}}{2},\dfrac{z-\bar{z_0}}{2i} \right)=0 \tag1 $
I don't "understand" this definition. Is there a way to see why the relation $(1)$ translates to our familiar geometric understanding of symmetry? Or alternatively, what does $(1)$ say about the points $z, z_0$ in different terms?
Note that $z=x+i y$ is symmetric to itself if and only if $F(x, y)=0$. That is, $z$ is symmetric to itself if and only if $z$ lies on the curve defined by $F$. Also since $F$ has real coefficients, the pair $(z,z_0)$ is symmetric if and only if $(z_0,z)$ is symmetric. Finally the map $z\mapsto z_0$ defined locally around a smooth point on $F$ which fixes $F$ pointwise is an anti holomorphic involution. (Anti holomorphic means $z\mapsto \overline{z_0}$ is holomorphic.) That makes this map by all accounts a very nice reflection!
If $F$ defines a line (or circle) then this map is the usual reflection (or inversion).
To conclude anti holomorphicity define the operators $$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$$ and $$\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$ as usual. Then $$0=\frac{\partial}{\partial \overline{z}}F\left(\frac{z+\overline{z_0}}{2}, \frac{z-\overline{z_0}}{2i}\right) = \frac{\partial F}{\partial \overline{z}}\left(\frac{z+\overline{z_0}}{2}, \frac{z-\overline{z_0}}{2i}\right) \cdot \overline{\frac{\partial z_0}{\partial z}}$$ and since $\frac{\partial F}{\partial \overline{z}}$ does not vanish near a smooth point on $F$ this implies $\frac{\partial z_0}{\partial z} = 0$ as claimed.