complex analysis question line I know that the map $$z\mapsto\frac{R^2}{\overline{z}-\overline{a}}+a$$ takes a point to a symmetric point respect to the circle $|z-a|=R$ and am trying to get the line $y=x$ in some form like $z=i\bar z$ but this is for $y=x$.
I am unsure of how to approach/start this problem.
Any help would be appreciated.
 A: When we have expressions in $x,y$ that we would like to have in $z$, we use the identities $Re(z)=\frac{z+\bar{z}}{2}$ and $Im(z)=\frac{z-\bar{z}}{2i}$. Your line is expressed by the equation
$$z-\bar{z}=3i(z+\bar{z})$$
or equivalently
Say that $z$ lies on that line. Then its symmetric point is given by
$$w=\frac{4}{\bar{z}-4}+4 $$
Solve for $\bar{z}$ and obtain $$\bar{z}=\frac{4}{w-4}+4$$
Replacing in the equation that is satisfied by $z,\bar{z}$ we get
$$\frac{1}{\bar{w}-4}-\frac{1}{w-4}=3i(\frac{1}{w-4}+\frac{1}{\bar{w}-4}+2) $$
Multiplying with both denominators
$$w-\bar{w}=3i(\bar{w}+w-8+2|w-4|^2)$$
using the trick about the real and the imaginary part again, this will yield an equation that I believe is going to be a circle. I believe you can work out the details from here!
EDIT: you changed the line from $y=3x$ to $y=x$. This changes the computations. Since $y=x$ your line is $z-\bar{z}=i(z+\bar{z})$.
The formula between $w,\bar{z}$ stays the same so
$$\frac{1}{\bar{w}-4}-\frac{1}{w-4}=i(\frac{1}{w-4}+\frac{1}{\bar{w}-4}+2) $$
Multiplying with both denominators
$$w-\bar{w}=i(\bar{w}+w-8+2|w-4|^2)$$
dividing both sides with $2i$
$$im(w)=re(w)-4+|w-4|^2 $$
writing $w=a+ib$
$$(a-4)^2+b^2-4+a-b=0$$
or
$$a^2-7a+12+b^2-b=0 $$
This is an equation of a circle. This is elementary analytic geometry, check out the internet for "analytic equation of a circle".
In the case of $y=3x$, the result was another circle. As mentioned in the comments, the result could be either a circle or a line. The method is the one I used above to get a 2nd degree analytic equation of the geometric set of points. With some elementary analytic geometry, you can then understand what this equation you get (in the way I explained) represents.
