I have to find the fixed points of the following linear transformation:
$$T(z) = \frac{(-\frac 13 -i\frac 23)z - \frac i2}{\frac z2 -1}$$ where $z \in \Bbb C$.
I think I have to solve the equation $T(z) = z$.
By placing $z = x+iy$ I arrived at the system of equations:
$$\begin{cases}3x^2-3y^2+2x+4y=0\\ 6xy+2y-4x-3=0 \end{cases}$$
I'm not sure I did it right