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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.

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You have made a good start and have been given a lot of hints in the comments. Basically, you just need to show the the stated properties are true, starting from the definitions of addition and multiplication of complex numbers:

$(a+bi)+(c+di)=(a+b)+(c+d)i$

$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$

where $a,b,c,d \in \mathbb R$. You can assume that addition and multiplication in $\mathbb R$ are well behaved.

Just be careful that you don’t use properties such as commutativity $x+y=y+x$ and associativity $(x+y)+z=x+(y+z)$ until you have proved that they are true in $\mathbb C$.

gandalf61
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