Consider a circle:
$C_R=\{w=(x,y): |w|^2=x^2+y^2=R^2\}$
Prove that $A(C_R)$ remains a circle if $A$ is either a conformal or an anticonformal matrix.
My attempt:
I defined the complex number $z:=x +iy$. Therefore, $\overline{z}=x-iy$.
$A(C_R)=A(|z|^2+|\overline{z}|^2)=A((Re^{it})^2+(Re^{-it})^2)=A(R^2)$
but it doesn't seem to take me anywhere!
One approach is to use a connection between complex numbers and $2\times 2$ matrices. This also helps in other situation. Namely, for a fixed complex number $a+ib$ the linear map $z\mapsto (a+ib)z$ can be written out as a linear transformation of $\mathbb R^2$ with the matrix $$\begin{pmatrix} a & -b \\ b & a\end{pmatrix} \tag 1$$ This is a conformal matrix. Conversely, any matrix of the form (1), that is antisymmetric and with constant diagonal, corresponds to the complex number $a+ib$, taken from its first column.
Similarly, anticonformal matrices come from linear maps $z\mapsto (a+ib)\overline{z}$:
$$\begin{pmatrix} a & b \\ b & -a\end{pmatrix} \tag 2$$
So, if a matrix of either form (1) or (2) is applied to the circle $|z|=R$, the image can be immediately written as the set of numbers $w=(a+ib)z$ (or $w=(a+ib)\bar z$) with $|z|=R$.