Let $\displaystyle f: (\mathbb{R}, +) \rightarrow (\mathbb{C}, \times )$ be the map $\displaystyle f(x) = e^{ix}$. Show that $\displaystyle f$ 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 $\displaystyle x,y\in (\mathbb{R}, +)$. Then $\displaystyle f(x+y) = e^{i(x+y)}=e^{ix + iy} = e^{ix}e^{iy} = f(x)f(y)$, so $\displaystyle f$ 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 $\displaystyle \{a\in (\mathbb{C}, \times ) : a = e^{ix}$ for some $\displaystyle x\in \mathbb{R}\}$.

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

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