Let $x.z=e^{ix}z$. Show that this is a group action by $\mathbb{R}$ on $\mathbb{C}$.
So far this is what I have got:
i) It has to satisfy $1_{\mathbb{R}}.z=z$
$1_{\mathbb{R}}=0$, so we get: $1_{\mathbb{R}}.z=e^{i*0}z=1*z=z$.
So i) is satisfied, now we need to satisfy:
ii) $x_1.(x_2.z)=(x_1x_2).z$
$x_1.(x_2.z)=x_1.e^{ix_2}z=e^{ix_1}e^{ix_2}z$
but $(x_1x_2).z=e^{ix_1x_2}z$
so ii) is not satisfied, even though it should be.
The semester is coming to a close so I might be missing something obvious here.. Any pointers?
Well, to define a group action... You need to have a group. In your question, you say Let $x$... omitting to say of which group you're speaking.
I imagine that the group is $(\mathbb R ,+)$. In which case, you indeed have a group action as
$$x_1 \cdot(x_2 \cdot z) = e^{ix_1}(e^{ix_2}z) = e^{i(x_1+x_2)}z= (x_1+x_2)\cdot z.$$