a). Prove that all the translations of the plane $R^2$ form a group under composition of functions (use vectors to write your proof).
b). Prove that all rotations of $C$ around a fixed point A form a group under composition of functions (use complex numbers to solve the problem).
For a) I have that if I let $u =<a,b>, v =<c,d>, w =<e,f> \in R^2$ and $(x,y) \in R^2$
Closure: $T_u \circ T_v (x,y) = (x+a+c, y+b+d) \in R^2$
Associative: Composition of functions is associative. I showed this to be true in an earlier problem.
Identity: If I let $e = <0,0>$ then $T_u \circ T_e (x,y) = T_u$ which implies that $T_e$ is the identity element.
Inverse: If I let $u =<a,b> , v =<-a,-b> $ then $T_u \circ T_v (x,y)= T_v \circ T_u (x,y) = T_e (x,y) = (x,y)$. So then $T_v = T_u ^ {-1}$.
So a) is true.
I'm having trouble showing that this is true for b). So far I have that if I let $P_{\theta}(z) = e^{i \theta}z$ and $v=a+bi$, then the rotation is $P(z) = e^{i \theta} z + v(1-e^{i \theta})$. So I have to show closure, associativity, that an identity exists, and an inverse exists. I am having trouble showing this is true under composition.
Any help would be appreciated.
Best Answer
To rotate the plane through an angle $\theta$ around a point $A$ you translate the plane by $-A$, rotate by $\theta$ and then translate by $A$ to put the plane back to where it was.
Thus, $r_{\theta}(z) = e^{i\theta}(z-A)+A$
Now,
$r_{\phi}(r_{\theta}(z)) = r_{\phi}(e^{i\theta}(z-A)+A) = e^{i\phi}((e^{i\theta}(z-A)+A)-A)+A= e^{i\phi}(e^{i\theta}(z-A))+A=e^{i(\phi+\theta)}(z-A)+A=r_{\phi + \theta}(z)$
Associativity follows by considering both $r_{\theta_1 + \theta_2}(r_{\theta_3}(z))$ and $r_{\theta_1}(r_{\theta_2 + \theta_3}(z))$.
$r_0$ serves as the identity and $r_{-\theta}$ is the inverse of $r_{\theta}$