There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$. Prove that, if $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.



Proposition: If $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

A (hypothesis): $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$.

B (conclusion): There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.



My Work

B1: $(a + bi)(c + di) = 1$

$\therefore a + bi \not = 0$

B2: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

A1: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$.

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A2: There is a complex number $x + yi$ such that $x + yi = \dfrac{1}{a + bi}$.

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$

A3: $(a+bi)(x + yi) = (a + bi)(c + di)$ where $a + bi \not = 0$

$\implies x + yi = c + di$ 

$Q.E.D.$



EDIT: My Work #2

A1: $(a + bi)(c + di) = 1$

$\implies ac + adi + bic + bdi^2 = 1$

$\implies ac + adi + bic - bd = 1$

$\implies (ac - bd) + i(ad + bc) = 1$ is a complex number, where $(ac - bd)$ is the real part and $i(ad + bc)$ is the imaginary part.

A2: Let $ac - bd = 1$ and $ad + bc = 0$

$\therefore 1 + 0 = 1$

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A3: Let $(a + bi)(x + yi) = 1$

A4: $(a + bi)(c + di) = (a + bi)(x + yi)$ where ($a \lor b) \not = 0$.

$\implies c + di = x + yi$ 

$Q.E.D.$



I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback. If there are any errors, please explain why and what the correct procedure is.
The aim of this exercise is to prove that the thing that you denote by $\frac1{a+bi}$ uniquely exists. Using this notation a priori implicitly assumes the statement we are about to prove.

Hint: Instead, as commented, try to explicitly find $c$ and $d$. For this, observe that $(a+bi)(a-bi)$ is real, and division is already known among reals.

For uniqueness of such $c+di$, we can take a general argument: 
Suppose $w_1$ and $w_2$ are both inverses of $z$, i.e. $w_1z=zw_2=1$.
Then we have
$$w_1 = w_1\cdot 1 = w_1zw_2=1\cdot w_2=w_2$$
It seem that both proofs are strange or incomplete.

$\dfrac1{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i$

This is a unique complex number.