Let $GL_2(\mathbb{C})$ the general linear group of order two on complex. We can define a action from $GL_2(\mathbb{C})$ on $\mathbb{C}^*$ as $$\begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}\cdot z=\displaystyle\frac{az+b}{cz+d} $$ prove that there exist only one orbit i.e for every $z\in \mathbb{C}^*$ we have $L_2(\mathbb{C})z=\mathbb{C}$.
I'd appreciate a hint to solve this problem. I think that for a given $z,w\in \mathbb{C}^*$ I should find a matrix $A\in GL_2(\mathbb{C})$ such that $A\cdot z =w$ but I don't know how.
Thanks!
You have exactly the right idea! The only question left is how one finds such a transformation.
Hint: If $z \neq 0$, we can always find a suitable transformation with $b = c = 0$