How to show that for all $a \in \mathbb{C}$ there is a unique $b \in \mathbb{C}$ such that $ab = 1$? Here is the whole question:
Show that for all $a \in \mathbb{C}$ with $a \neq 0$, there exists a unique $b \in \mathbb{C}$ such that $ab = 1$. 
I believe I have shown that it exists.
Proof:
Suppose $a \in \mathbb{C}$ where $a \neq 0$. Let $b = 1/a \in \mathbb{C}$. Then $ab = a(1/a) = a/a = 1$. Thus there exists a $b \in \mathbb{C}$ such that ab = 1.
I am confused on how to show that it is unique. Do I suppose that there exist a $b_1$ and $b_2 \in \mathbb{C}$ such that $ab_1 = 1$ and $ab_2 = 1$? If I do that, I have $ab_1 = ab_2 = 1$, then I can multiply on the left by the multiplicative inverse of $a$. That gives me $b_1 = b_2 = 1/a$. Is that okay? It seems kind of circular to me, since earlier I took $b = 1/a$.
Also, I found a solution online for showing uniqueness, but I don't understand the mechanics of how this is showing that it is unique:
If $ab = 1$, then
b = $1 \cdot b$ = $(\frac{1}{a} \cdot a) \cdot b = \frac{1}{a} \cdot (a \cdot b) = (\frac{1}{a} \cdot 1) = \frac{1}{a}$. 
 A: You need to be careful. What do you mean by $1/a$? It might sound like I'm being silly, because you want to say "of course I know what I mean by $1/a$; it's $1$ divided by $a$". But what do you mean by dividing by $a$? If I write down the symbols $$\frac{1}{1+i}$$
what does that mean? I'm claiming (well, you are) that it's a complex number. Well, if it's a complex number, what is its real part? What is its imaginary part? It's not clear.  
Alright, now that's out of the way, let's find our $b$. We really hope that our usual properties for $\mathbb{R}$ hold, so we would hope that $$\frac{1}{1+i}=\frac{1-i}{(1+i)(1-i)}=\frac{1}{2}(1-i)$$
Just to be clear, the first thing we wrote might not be a complex number, it's just a meaningless collection of symbols at the moment. But the last term is definitely a complex number! We absolutely know how to divide by real numbers like $2$, so we're fine. Now for the real test, is it really an inverse for $1+i$? Well, $$(1+i)\frac{1}{2}(1-i)=2\cdot \frac{1}{2}=1$$
Exactly as we hoped! In fact, for any complex number $a$ we have $$a\cdot \frac{\bar{a}}{|a|^{2}}=1$$
and therefore we define the collection of symbols $$\frac{1}{a}=\frac{1}{|a|^{2}}\bar{a}$$
And this is a bonafide complex number that is an inverse for $a$. Luckily, this satisfies all the usual properties of fractions over $\mathbb{R}$ that I won't talk about in detail.  
Finally, we want to show it's unique. But that's not too bad: Suppose $ab=1$ and $ac=1$. Then $ab-ac=a(b-c)=0$. Now multiply by $b$ (or $c$, it doesn't matter), to cancel the $a$, and since $b\cdot 0=0$, we get $b-c=0$. So $b=c$! So there must be only one inverse of $a$. And we're done.
A: You claim you know that $\mathbb C$ is a field with a multiplicative identity, 1, and the property that for every $a \ne 0$ there exists a unique multiplicative inverse, 1/a, such that 1/a*a = 1.
We must prove that by identifying what the value of 1/a $\in \mathbb C$ is, and be showing it is a unique solution to $ab = 1$.
=== old =  skip the first two paragraphs ===
You need to specify that $a \ne 0$.
You can't claim $1/a$ is the value because you do not know that $1/a$ exists and is well defined.  That $1/a$ exists is well-defined and exist is precisely what you are being asked to prove.
Basically we know that $1/a$ exist because $\mathbb C$ is a field and that is part of the definition of field.  However we have to prove $\mathbb C$ is a field.  Or in this case just that $\mathbb C$ satisfies the inverse axiom of fields.
==== okay, start reading from here ====
Let's let $a = x + i y \in \mathbb C; a \ne 0$.  If we can solve for $b = w + i z$ such that $ab =1$ and that $b$ is a unique solution, we are done.
$(x + iy)(w + iz) = 1$
$(xw - yz) + i(wy + xz) = 1$
$xw - yz = 1; wy + xz = 0$
$wy = -xz$
Case 1: $x = 0$
Then $y \ne 0$ (as $x + iy \ne 0)
So $wy = 0$. So $w = 0$
$xw -yz = 1$
$-yz = 1$
$z = -1/y$.
There was no ambiguity so $b = (-1/y)i = (x - i y)/(x^2 + y^2)$ is the unique solution to $iy*b = 1; y \ne 0; x = 0$.
Case 2: $y = 0$.  In this case $a = x \in \mathbb R$ and $b = 1/x = (x - i y)/(x^2 + y^2)$ is unique solution.
Case 3: $x \ne 0; y \ne 0$
$xw - yz = 1; wy + xz = 0$
$wy = -xz$
$w = -xz/y$
$-x^2z/y - yz = 1$
$z(x^2/y + y) = -1$
Note: $(x^2/y + y) = 0 \implies x^2 = -y^2$ which is impossible so $(x^2/y + y) \ne 0$.
$z = -1/(x^2/y + y) = -y/(x^2 + y^2)$
$w = -xz/y = x/(x^2 + y^2)$ 
So $b = (x - y i)/(x^2 + y^2)$ is the unique solution to $(x + y i)*b = 1; x + yi \ne 0$.
A: When it comes to basic properties like this (we take existence and uniqueness of multiplicative inverse for granted, more often than not!) it really is important to be pedantic and dot every i. Otherwise, you might as well just say "Well, everyone knows that, of course it's true."
OK, so following Axler's text, all we know is that complex numbers look like $a + bi$, where $a$ and $b$ are real numbers, and they have the addition and multiplication that we all know. So, if you want to show that some nonzero $z = a + bi \in \Bbb C$ has an inverse, it really isn't good enough (in my opinion) to say that $1/z$ works. Does $1/z$ look like $a + bi$ for some real $a, b$? If you can't show how to put it in that form, then you don't know $1/z$ is indeed a complex number.
Now, suppose you've spelled out how to write $1/z$ in the $a + bi$ form (I do hope you remember how that works; use the conjugate of the denominator). Also, let's suppose that you've proven the other properties about $\Bbb C$ that we all know, namely

*

*multiplication and addition are commutative and associative (truly grueling to demonstrate) and that


*we have additive and multiplicative identities.
We'd like to show that that the multiplicative inverse of a fixed but arbitrary nonzero $z \in \Bbb C$ is unique.
The standard computation that you're likely to see in a text on group theory is the following. It relies on the existence of inverses, the existence of the (multiplicative) identity $1$, and associativity.

Given $0 \neq z \in \Bbb C$, suppose $w_1$ and $w_2$ are both multiplicative inverses of $z$. Among other things, this means that $z w_2 = 1 = w_1 z$. Then we have
$$w_1 = w_1 \cdot 1 = w_1(z w_2) = (w_1 z)w_2 = 1 \cdot w_2 = w_2,$$
so that $w_1 = w_2$ and the inverse of $z$ is indeed unique, as desired.

Since we know so little about $\Bbb C$ from the axioms given, it's mostly just a trick and not terribly interesting -- but make sure you can say exactly what property is used at each step. The same kind of computation, replacing $\cdot$ with $+$, shows that additive inverses are unique too, assuming the corresponding properties of addition
A: $1/a$ is just a notation. To prove the existence of b, write $a=c+id$ and so $b=\frac{c-id}{c^{2}+d^{2}}$ is the inverse of $a$. Indeed $$ab=(c+id)(\frac{c-id}{c^{2}+d^{2}})=\frac{c^{2}+d^{2}}{c^{2}+d^{2}}=1$$
and, similarly, $ba=1$.
If $b_{1}$ is such that $ab_{1}=b_{1}a=1$, then $$b=b1=b(ab_{1})=(ba)b_{1}=1b_{1}=b_{1}$$.
