Prove that the following functions are group homomorphisms, give their kernel and range and show if they are injective or surjective. For each proof we show that $f(a\circ b)=f(a)\circ f(b)$. My question here is if the proofs are correct, and whether they could be done in a more efficient manner for my own improvement, as well as having examples for other readers.
(1) $f:(\mathbb{R},+)\to (\mathbb{C},*)$ defined by $f(x)=e^{2\pi ix}$
$f(a+b)=e^{2\pi i(a+b)}=e^{2\pi ia+2\pi ib}=e^{2\pi ia}*e^{2\pi ib}=f(a)*f(b).$
So $f$ is a group homomorphism.
$N(f)= \{x\in\mathbb{R}: f(x)=1\}$
We solve for $e^{2\pi ix} = \cos(2\pi x)+i\sin(2\pi x) = 1$. This holds if and only if:
$\cos(2\pi x)=1 \land i\sin(2\pi x)=0$ which holds for $x=k,k\in\mathbb{Z}.$
As $N(f)=\mathbb{Z}\neq\{1\},$ the function is not injective.
As the outputs of $\cos(x)$ and $\sin(x)$ are in between $0$ and $1$:
$R(f)=\{a+bi\in\mathbb{C}: -1\leq a\leq1 \land -1\leq b\leq 1\}\neq(\mathbb{C},*)$ so the function is not surjective.
(2) $f:(\mathbb{C},*)\to(\mathbb{R},*)$ defined by $f(z)=z\bar{z}$
$f((a+bi)(c+di))=f((ac-bd)+(ad+bc)i)=a^2c^2+b^2d^2+b^2c^2+a^2d^2 =$ $(a^2+b^2)(c^2+d^2)=(a+bi)(a-bi)*(c+di)(c-di)=f(a+bi)*f(c+di)$.
So $f$ is a group homomorphism.
$N(f)=\{a+bi\in\mathbb{C}:a^2+b^2=1\}$
Take $x_1=1+i\in\mathbb{C}: f(x_1)=1^2+1^2=2$
Take $x_2=-1-i\in\mathbb{C}: f(x_2)=(-1)^2+(-1)^2=2$
We have $x_1\neq x_2$ but $f(x_1)=f(x_2)$ so the function is not injective.
$R(f)=\{x\in\mathbb{R}:a^2+b^2=x$ with $a,b\in\mathbb{R}\}$
As $\forall_{a,b\in\mathbb{R}}: a^2+b^2\geq 0$, we have that $-1\notin R(f)$ but $-1\in(\mathbb{R},*)$.
So, $R(f)\neq(\mathbb{R},*)$ so $f$ is not surjective.
(3) $f:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $f(a)\to\bar{a}$
$f(a+b)=\bar{a+b}=\bar{a}+\bar{b}=f(a)+f(b)$ so $f$ is a group homomorphism.
$N(f)=\{a\in\mathbb{Z}: f(a)=0\}$
We have $n\in\mathbb{Z}: f(n)=0$ and $2n\in\mathbb{Z}: f(2n)=0$
This way, $N(f)\neq\{0\}$ so $f$ is not injective.
$R(f)=\{\bar{a}\in\mathbb{Z}: f(a)=\bar{a}$ with $ a\in\mathbb{Z}\}$
We know that for all functions $f:A\to B$ that $R(f)=f(A)\subset B$.
In this case: $\mathbb{Z}/n\mathbb{Z} \supset f(\mathbb{Z})\supset f(\{0,1,2\dots n-1\})=\{0,1,2\dots n-1\}$
As $\mathbb{Z}/n\mathbb{Z} = \{0,1,2\dots n-1\} $ we have $\{0,1,2\dots n-1\}\subset f(\mathbb{Z})$ and $f(\mathbb{Z})\subset \{0,1,2\dots n-1\} $ so that $R(f) = \mathbb{Z}/n\mathbb{Z} $. Thus, $f$ is surjective.