I was given a task to prove the following properties of $\mathbb {C}$
identities
$x + 0 = x $ and $ x1 = x$ for all $x\in\mathbb {C} $
additive inverse
$\forall a \in \mathbb {C}$ there exists a unique $b \in \mathbb {C}$ such that $a + b = 0$
distributive property
$x(a + b) = xa + xb$ for all $x, a, b \in \mathbb {C}$
Since I am a novice in mathematics, I have no clue how can I prove that. Please provide some clarification.
Proof on identities.
Suppose $x = a + bi$ where $a, b \in \mathbb {R}$ and $0 = 0 + 0i$. Therefore $a + bi + 0 + 0i = a + bi$, using the operation addition defined on $\mathbb {C}$. Using the same argument we can prove that $x1 = x$.
Proof on additive inverse.
Suppose $z = a + bi$ is an arbitrary complex number and let $w = -a + -bi$. Then
$z + w = (a + (-a)) + (b +(-b))i = 0 + 0i = 0$ That is $w$ is the additive inverse of z. Let's show that $w$ is unqiue.
Suppose that there exist $y : z + y = 0$ and $y \neq w$.
Then $z + w = 0$
$z + w + y = y$
$(z + y) + w = y$
$ 0 + w = y $ Contradiction. Therefore w is unique.