Let $\theta: \Bbb C^∗ \to \Bbb R_+$ by $\theta (z)=|z|$

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How to show $\theta$ is a homomorphism from $(\Bbb C, \times)$ onto $(\Bbb R_+, \times)$.

Let $\theta: \Bbb C^∗ \to \Bbb R_+$ by $\theta (z)=|z|$

What I have so far is:

$$\theta(ab) = |ab|$$

$$\theta(a) + \theta(b) = |a||b| = |ab|$$

does this mean that they are homomorphism?

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Let $(G, \circ)$ and $(H, \diamond)$ be groups. A homomorphism is a map $\theta: G \rightarrow H$ that preserves the group operations. I.e. for $g, h \in G$

$$\theta(g \circ h) = \theta(g) \diamond \theta(h).$$

In your case, $G = \mathbb{C}^{*}$ and $H = \mathbb{R}^+$ with the operations of multiplication in both cases.

Let $\theta(z) = \lvert z \rvert$ with $z \in \mathbb{C}^*$. To check if this is a homomorphism, we simply apply the definition. Let $z_1$ and $z_2$ be elements of $\mathbb{C}^*$. Then,

$$\theta(z_1z_2) = \lvert z_1 z_2\rvert = \lvert z_1\rvert \lvert z_2 \rvert = \theta(z_1)\theta(z_2),$$

as was to be shown.