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?

$\endgroup$
2
  • 3
    $\begingroup$ There is this guy called DeMoivre. $\endgroup$
    – Pedro
    Commented Mar 6, 2014 at 2:57
  • 1
    $\begingroup$ $\mathbb R$ is a group for $+$, not $\times$! $\endgroup$
    – mookid
    Commented Mar 6, 2014 at 3:34

2 Answers 2

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
  • 1
    $\begingroup$ The point is sometimes we define $$e^{a+ib}=e^a(\cos b+i\sin b)$$ $\endgroup$
    – Pedro
    Commented Mar 6, 2014 at 3:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .