0
$\begingroup$

Let $\phi :\mathbb{R}\rightarrow \mathbb{S}$ be defined by

$\hspace{150pt} \phi(r)=e^{2\pi ri}=\cos(2\pi ri)+i\sin(2\pi ri)$

Show that this is a group homomorphism.

$\mathbb{S}$ is the unit circle.


If I consider two elements $q,r\in\mathbb{R}$ then

$\hspace{100pt} \phi(qr)=e^{2\pi(qr)i}=\cos(2\pi (qr)i)+i\sin(2\pi (qr)i)$

and

$\hspace{50pt} \phi(q)\phi(r)=e^{2\pi qi}e^{2\pi ri}=(\cos(2\pi qi)+i\sin(2\pi qi))(\cos(2\pi ri)+i\sin(2\pi ri))$

I haven't been able to make $\phi(qr)=\phi(q)\phi(r)$ though. Am I missing an identity or something?

$\endgroup$
  • 3
    $\begingroup$ There is this guy called DeMoivre. $\endgroup$ – Pedro Tamaroff Mar 6 '14 at 2:57
  • 1
    $\begingroup$ $\mathbb R$ is a group for $+$, not $\times$! $\endgroup$ – mookid Mar 6 '14 at 3:34
1
$\begingroup$

Remember that groups have operations with associative laws. You did not note that $\mathbb{R}$ is the additive group of reals. Now you can take $a,b\in \mathbb{R}$ with $a+b \in \mathbb{R}$ since groups are closed. Then $\phi(a+b)=e^{(a+b)(2\pi i)}=e^{2\pi ai}e^{2\pi bi}=\phi(a)\phi(b)$ and so $\phi$ is a homomorphism.

$\endgroup$
0
$\begingroup$

There's no need to use cosine and sine simply note: $e^{2\pi qi}e^{2\pi r i}=e^{2\pi i(q+r)}$.

$\endgroup$
  • 1
    $\begingroup$ The point is sometimes we define $$e^{a+ib}=e^a(\cos b+i\sin b)$$ $\endgroup$ – Pedro Tamaroff Mar 6 '14 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.