Proof: For every $\alpha \in \mathbb{C}$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha + \beta = 0$

Question

I am trying to prove the following:

For every $\alpha \in \mathbb{C}$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha + \beta = 0$.

There are two steps to this proof:

1. I first need to prove that, for every $\alpha \in \mathbb{C}$, there does exist $\beta \in \mathbb{C}$ such that $\alpha + \beta = 0$.
2. I then prove that, for every $\alpha \in \mathbb{C}$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha + \beta = 0$.

My proof follows.

Let $\alpha \in \mathbb{C}$ and $\beta \in \mathbb{C}$.

1. $$\begin{align} \alpha + \beta &= (a + bi) + (c + di), \ \text{where $a, b, c, d \in \mathbb{R}$} \\ &= (a + c) + (b + d)i \ \ \ \text{(By the definition of complex addition.)} \\ &= (a + (-a)) + (b + (-b))i \\ &\text{(Since $c, d \in \mathbb{R}$ and $a, b \in \mathbb{R}$,} \\ &\text{which implies that $-a, -b \in \mathbb{R}$, and so we set $c = -a$ and $d = -b$.)} \\ &= (0) + (0)i \\ &= 0 \tag*{$\blacksquare$} \end{align}$$

2. Here, I prove that $\beta$ is unique by assuming that there are two objects $\beta_1$ and $\beta_2$, and then show that they must therefore be the same object.

$$\begin{align} \alpha + \beta_1 = 0, \\ \alpha + \beta_2 = 0 \end{align}$$

Subtracting the two equations, we get

$$\begin{align} &(\alpha + \beta_1) - (\alpha + \beta_2) = 0 - 0 \\ &\Rightarrow [(a + bi) + (c_1 + d_1 i)] - [(a + bi) + (c_2 + d_2 i)] = 0, \text{where $a, b, c_1, c_2, d_1, d_2 \in \mathbb{R}$} \\ &\Rightarrow [(a + c_1) + (b + d_1)i] - [(a + c_2) + (b + d_2)i] = 0 \ \ \ \text{(By the definition of complex addition.)} \\ &\Rightarrow [(a + c_1) - (a + c_2)] + [(b + d_1) - (b + d_2)]i = 0 \ \ \ \text{(By the definition of complex addition.)} \\ &\Rightarrow (a + c_1 - a - c_2) + (b + d_1 - b - d_2)i = 0 \ \ \ \text{(By distributivity.)} \\ &\Rightarrow (a - a + c_1 - c_2) + (b - b + d_1 - d_2)i = 0 \ \ \ \text{(By commutativity.)} \\ &\Rightarrow (c_1 - c_2) + (d_1 - d_2)i = 0 \\ &\Rightarrow c_1 - c_2 + d_1 i - d_2 i = 0 \\ &\Rightarrow c_1 + d_1 i = c_2 + d_2 i \\ &\Rightarrow \beta_1 = \beta_2 \tag*{$\blacksquare$} \end{align}$$

I would greatly appreciate it if people could please take the time to review my proof. If there are any errors, please point out the specific error and explain the correct way.