Homomorphism, kernal, image

**Pinkk** · February 8th 2011, 03:53 PM

Let $f: (\mathbb{R}, +) \rightarrow (\mathbb{C}, \times )$ be the map $f(x) = e^{ix}$ . Show that $f$ is a homomorphism and determine its image and kernal.

Okay, so this seems somewhat easy, but I am fairly new to this concept and groups in general, so I want to check if my work is correct.

Proof: Let $x,y\in (\mathbb{R}, +)$ . Then $f(x+y) = e^{i(x+y)}=e^{ix + iy} = e^{ix}e^{iy} = f(x)f(y)$ , so $f$ is a homomorphism. Q.E.D.

Am I allowed to distribute in the exponent? I am not sure since we are talking about groups with different laws of composition and no mentioning of distribution or anything, so any feedback on that would be appreciated.

As for the image, I think it's just the set $\{a\in (\mathbb{C}, \times ) : a = e^{ix}$ for some $x\in \mathbb{R}\}$ .

As for the kernel, since the identity of $(\mathbb{C}, \times )$ is 1, we have that $f(x) = 1$ whenever $ix = 0$ , or when $x=0$ , so the kernel of $f$ is $\{0\}$ .

Again, any feedback/critique would be appreciated, thanks.

**LoblawsLawBlog** · February 8th 2011, 04:12 PM

Your proof of the homomorphism property is fine. Your description of the image is also correct but you should be able to describe it as something familiar.

There's more to the kernel.
$f(x)=1\Rightarrow e^{ix}=1$ .
For real exponents, $e^x=1\Leftrightarrow x=0$ , but this isn't true for complex numbers.

**DrSteve** · February 8th 2011, 04:41 PM

One more hint for the kernel:

$e^{2\pi i}=1$

**Pinkk** · February 8th 2011, 04:52 PM

I don't recognize how I could describe the image any other way.

And thanks, I haven't dealt with complex numbers in a long time (real analysis took up my last year of math learning).

**DrSteve** · February 8th 2011, 04:56 PM

The image is just the unit circle $|z|=1$

The kernel consists all integer multiples of $2\pi$

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