Proof: For every $\alpha \in \mathbb{C}$ with $\alpha \not= 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$ I am trying to prove the following:

For every $\alpha \in \mathbb{C}$ with $\alpha \not= 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$.

There are two steps to this proof: 


*

*I first need to prove that, for every $\alpha \in \mathbb{C}$, where $\alpha \not= 0$, there does exist $\beta \in \mathbb{C}$ such that $\alpha\beta = 1$.

*I then prove that, for every $\alpha \in \mathbb{C}$, where $\alpha \not= 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha\beta = 1$.


My proof follows.
Let $\alpha \in \mathbb{C}$, where $\alpha \not= 0$, and $\beta \in \mathbb{C}$. Also, I will assume that it is valid to assume knowledge of the fact that the inverse of a complex number is also a complex number. Otherwise, I'm not sure how one would do the proof.


*

*\begin{alignat}{2}
\alpha\beta &= \alpha \alpha^{-1} & &\text{(The inverse of a complex number is also a complex number.} \\
& & &\text{Therefore, since $\alpha \in \mathbb{C}$ and $\beta \in \mathbb{C}$, we can set $\beta = \alpha^{-1}.)$} \\
&= 1 \tag*{$\blacksquare$}
\end{alignat}

*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.
$$\begin{align} \alpha \beta_1 = 1, \\ \alpha \beta_2 = 1 \end{align}$$
Subtracting the two equations, we get 
$$\begin{align} &\alpha \beta_1 - \alpha \beta_2 = 0 \\ &\Rightarrow \alpha(\beta_1 - \beta_2) = 0, \\ &\Rightarrow \beta_1 - \beta_2 = 0 \ \ \ \text{(Since $\alpha \not= 0$.)} \\ &\Rightarrow \beta_1 = \beta_2 \ \ \ \tag*{$\blacksquare$} \end{align}$$
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.
 A: With $$\alpha=x+iy,\beta=a+bi$$ we get
$$ax-by+i(ay+bx)=1$$ so you have to solve
$$ax-by=1$$ and $$ay+bx=0$$
solving this system you will get
$$(a^2+b^2)x=a$$ if $$a\neq 0$$ so $$x=\frac{a}{a^2+b^2}$$
A: Your proof assumes that each complex number has an inverse. You are supposed to prove that. And how do you know that $\alpha(\beta_1-\beta_2)=0$ and $\alpha\neq0$ implies that $\beta_1-\beta_2=0$?
A: Given $\alpha = a_R+a_Ii$ in $\mathbb C$ with $a_R, a_I \in \mathbb R$ and $a_R$ or $a_I$ non-zero, let $\beta = b_R+b_Ii, $ where $$b_R=\frac {a_R}{a_R^2+a_I^2}$$ and
$$b_I=\frac {-a_I}{a_R^2+a_I^2}.$$
Since $a_R$ or $a_I$ is non-zero, $a_R^2+a_I^2$ is non-zero, so $b_R$ and $b_I$ exist as real numbers and $\beta \in \mathbb C$.  
You can show that $\alpha \beta = 1$.
A: As José Carlos Santos points out in his answer, the fact that every $\alpha \in \mathbb{C}$ with $\alpha \ne 0$ has an inverse is precisely what you have to prove, so you can't assume it in your proof.
J. W. Tanner's answer is technically correct, but it does rather assume that you already "magically" know what the inverse $\beta$ of $\alpha$ is.
In contrast, Dr. Sonnhard Graubner's answer shows how you could "discover" for yourself that the inverse $\beta$ of $\alpha$ exists, and that it has the form given in J. W. Tanner's answer.
However, I find some aspects of his answer confusing, so - with apologies - I present a slightly different version of essentially the same argument. (I hope that I haven't just introduced confusions of my own!)
First, I make some further remarks on the proof you've presented:

One flaw in your proof that hasn't been mentioned is that you repeat essentially the same argument in 1 and 2.
In 1, you conclude from $\alpha\beta = \alpha\alpha^{-1}$ that $\beta = \alpha^{-1}$, but you don't give a reason why.
In 2, you conclude from $\alpha\beta_1 = \alpha\beta_2$ that $\beta_1 = \beta_2$, and this time you do give a reason. However, as José Carlos Santos points out, you give no justification for concluding that if $\alpha(\beta_1 - \beta_2) = 0$ then $\beta_1 - \beta_2 = 0$. The only obvious way to justify this is by multiplying both sides by an inverse of $\alpha$. In 2 (unlike 1), you can do this without logical circularity. Still, the proof is unnecessarily complicated.

Let $\alpha = x + iy$. We are given that $\alpha \ne 0$, i.e. that $x \ne 0$ or $y \ne 0$.
Let $\beta = u + iv$: then $\alpha\beta = (xu - yv) + i(xv + yu)$.
The equation $\alpha\beta = 1$ is therefore equivalent to two simultaneous linear equations in real variables $u$ and $v$:
\begin{align*}
xu - yv & = 1, \\
xv + yu & = 0.
\end{align*}
It is up to you now to solve these equations for $u$ and $v$, with the help of the condition that $x \ne 0$ or $y \ne 0$.
The fact that a solution exists means that an inverse $\beta$ of $\alpha$ exists.
The fact that the solution is unique means that the inverse $\beta$ is unique - and now one can denote it by $\alpha^{-1}$.
