Why the map sending $a\otimes b$ to $(ab, a\bar{b})$ is injective?

Question

I am wondering why the ring homomorphism $\phi : \mathbb{C}\otimes_\mathbb{R}\mathbb{C} \longrightarrow \mathbb{C}\times\mathbb{C}$ sending $a\otimes b$ to $(ab, a\bar{b})$ is one-to-one. So any suggestion would be helpful.

Answer 1

Hint: A ring homomorphism is injective if and only if it has trivial kernel. For which $a, b \in \mathbb{C}$ is $ab = 0$ and $a\bar{b} = 0$?