What are the fixed points of this action?

Question

For a fixed integer $d$, let $G=\left\{\left(t,\dfrac{1}{t}\right)\in (\mathbb C^*)^2:t^d=1\right\}$ act on $\mathbb C^2$ by ponitwise multiplication. That is $$\left(t,\dfrac{1}{t}\right)\cdot(x,y)=\left(tx,\dfrac{1}{t}y\right)$$

I need help finding the fixed points of this action.

Obviously $(0,0)$ is a fixed point. But are there any others?

My attempt :

If $\left(t,\dfrac{1}{t}\right)\cdot (x,y)=(x,y)\Longrightarrow tx=x\quad\text{and}\quad ty=y$.

Now writing $t=e^{\frac{2\pi ik}{d}}\quad\text{where}\quad 0\leq k\leq d-1$, $x=r_1e^{it_1}$ and $y=r_2e^{it_2}$ we have, $$r_1e^{it_1+\frac{2\pi ik}{d}}=r_1e^{it_1} \quad\text{and}\quad r_2e^{it_2+\frac{2\pi ik}{d}}=r_2e^{it_2}$$ Suppose $r_1\neq0$ we get $$e^{it_1+\frac{2\pi ik}{d}}=e^{it_1}\Longrightarrow it_1+\dfrac{2\pi ik}{d} =it_1 + 2\pi i n \text{ for some }n\in\mathbb Z\Longrightarrow k=dn$$ But since $0\leq k\leq d-1$ this cannot happen. Similarly for $r_2\neq 0$. Thus the only fixed point is $(0,0)$.

Is everything I have done correct?

Thank you.

Answer 1

Much easier: $(x,y)\ne(0,0)\implies$ $x\ne 0$ or $y\ne 0$ $\implies$ $t=1$ because $tx=x$ and $ty=y$.

Answer 2

Your argument looks correct, but really it's much simpler: unless $d=1$, there is at least one $t \neq 1$ such that $t^d = 1$, and then $(t, 1/t) \cdot (x,y) = (x,y) \implies tx = x \text{ and } ty = y$. But since $t \neq 1$, $tx = x \implies (1-t)x = 0 \implies x = 0$, and similarly $y = 0$.

(Of course if $d=1$ then $G$ is the trivial group, so everything is a fixed point.)