Let $G=(\mathbb C^*,\cdot), G'=(\mathbb R^*,\cdot)$ and $\phi : G\to G'$ be defined by $\phi(z)=|z|, z\in \mathbf C^*$, where $\mathbf C^*=\mathbf C-\{0\}$ and $\mathbf R^+$ is a set of all positive real numbers. Then we have to
show that $\phi$ is a homomorphism.
determine $\ker \phi$.
determine $Im \phi$
Solution :
Let $x,y\in G$. Then $\phi(x)=|x|, \phi(y)=|y|$.
Now $\phi(xy)=|xy|=|x||y|=\phi(x)\phi(y)$
Hence $\phi$ is a homomorphism.
For the second and third,
$\ker\phi=\{x\in G: \phi(x)=0\}$ and $Im \phi=\{ \phi(x)\in G': x\in G\}$. How can I determine $\ker\phi$ and $Im \phi$.
The identity element in $\Bbb R^*$ is $1$ and not $0$ so the kernel is $$ \{x\in G: \phi(x)=1\} = \{z \in \Bbb C \setminus \{0\} \ \mid \ |z| = 1 \} $$
This set can either be described as all elements on the unit circle on the complex plane or as the set of all $\{\cos \theta + i \sin \theta \ \mid \ \theta \in \Bbb R\}$.
As for the image of $\phi$ notice that any positve element in $ \Bbb R \setminus \{0\} $ is also an element in $ \Bbb C \setminus \{0\} $ and so you should be able to see that this map is onto $(0 , \infty)$.