Why are complex numbers defined as $a+bi$? I understand that a complex number $n = a + bi$ is defined as having real $a,b$ with $i = \sqrt{-1}$. However what I don't understand is the why. Why was it defined this way? How do we know this will be a useful way to define them? Was this emergent from previous mathematics or a defined "workaround" to address certain problems? Like what led to someone going, "Hmm, we should define a new type of number and represent it as $a + bi$ where $i = \sqrt{-1}$!" and so on.
I have been trying to understand the basics of complex analysis but I haven't yet understood why numbers are represented this way before I can wrap my head around why all these other interesting areas of mathematics work themselves out. Like I could easily envision myself defining a specific number in some way only to find later that it was a bad definition or an incomplete or inaccurate way to describe something. What makes $a + bi$ correct and why did it come about?
 A: Historically, complex numbers were introduced by Cardan as far back as the 16th century. While people did not really understand them at the time (much like undergrads now!), introducing this famous notation $a+b\sqrt{-1}$ served in intermediate computations for finding explicit solutions to cubic equations (with real solutions!).
So if it Cardan's intuitive idea could be made rigorous (and it can, of course), then it at least has the merit of allowing one to do computations with more general numbers than just reals, before returning to reals. This is by the way also useful for factoring polynomials like $X^4+1$ over the reals.
Of course, by now we (mathematicians, physicists, engineers...) are so used to the abstract notion of complex numbers that we think of them as just as natural as reals or integers. Also, it turns out that they have this wonderful property that every polynomial (real or complex) can be factored into a product of complex polynomials of degree one. This means that complex numbers are enough to study any polynomial with real coefficients.
A: You start with the question "What is $\sqrt {-1}$?" 
And you answer that by creating a value for it that isn't a "real number", called $i$, and create a new number system that is linear combinations of $1$ and $i$ -
 that works. And gives sensible answers to other square roots, other polynomial roots, and has consistent algebra that goes beyond real number results, and then has natural (if sometimes unexpected) extensions of existing functions, etc.
If it didn't produce a set of sensible consistent results, we probably would never have heard of it.
A: The definition is "right" because it makes it into a complete minimal field extension of the real numbers that makes every polynomial with real coefficients have a solution.
A: This might not be the exact reason why they are defined this way (there might be many reasons, in fact), but it is sometimes useful to think of complex numbers as objects that are analogous to 2 dimensional vectors. We could define:
$f:\mathbb{C}\rightarrow\mathbb{R}^2$
such that $f(z)=(a,b)$ where $z=a+bi\in\mathbb{C}$ and $(a,b)\in\mathbb{R}^2$.
Hence the real and imaginary parts play the role of cartesian coordinates.
It is also a well-known fact that every complex numbers can be rewritten in polar form such that $z=re^{i\theta}$, where $r$ is the modulus of the vector and $\theta$ its orientation. In fact this form is another way to define complex numbers. 
A: $C$ is the algebraic closure of $R$. That is all polynomials with real coefficients(even complex ones) split in $C[x]$. And $C$ is the smallest extension of $R$ with this property.
A: Two questions
I think there are two main parts to this question:

*

*Why were complex numbers first used, and what makes them useful now?

*What makes $a+bi$ a useful/good representation?

1. Why are complex numbers useful?
This is addressed well in Glougloubarbaki's answer. Briefly, they were first conceived because when people were first solving cubic equations (with real coefficients) for real roots, they noticed that calculations involved square roots of negative numbers arising, but that things worked out if you just assumed that those were okay. It was later shown that these "imaginary roots" are unavoidable if you want to write down the roots of a cubic using radicals (and not trig functions or something); a key phrase is "casus irreducibilis".
Now there are lots of reasons that complex numbers are used in different contexts, but I'd bet the vast majority of them are either related to nice rotation-related properties beyond the scope of this answer (e.g. "phasors" in electrical engineering) or to the Fundamental Theorem of Algebra, which says that expanding the reals to the complex numbers is enough if you want all the polynomials to have as many roots as they should.

2. Why $a+bi$?
I think this is an excellent question that doesn't get explicitly discussed enough; this is by no means obvious. If you instead wanted to, say, add on a cube root of $2$ to the rationals (call it $j$), then "all numbers of the form $a+bj$" would not be good at all.
In answering, I want to directly address the comment "...what lets us add, subtract, and multiply them? What lets us treat them as vector quantity?". For what follows, I will suppose that $i$ satisfies $i^2=1$ and that we want to be able to do arithmetic with our complex numbers and follow the usual rules like commutativity of multiplication.
Why do we need at least $a+bi$?
Before we show that the numbers of the form $a+bi$ are good enough, we should check that we need all of them. For instance, maybe we only need nonnegative values of $b$ or something.
Pure imaginaries
Well, since we want to be able to multiply numbers, we need numbers of the form $bi$ since that would be a product of a real number $b$ and the new number $i$.
Why are all these products distinct/necessary? Well, $(bi)^2=b^2i^2=-b^2$, so for every nonnegative number $b$, these numbers have to be different. But that doesn't rule out something like $3i=-3i$ being true. But $3i+(-3i)=0$ and $3i+3i$ squares to $-36$, so they can't be equal. Similarly for any other real number in place of $3$.
Adding in $a$.
Then we could add a real number to these new "pure imaginary" numbers like $bi$, to get numbers like $a+bi$. But again we should check that we really need all of these.
If $a+bi=c+di$, then $a-c=di-bi$. Squaring both sides, we find that $(a-c)^2=-(d-b)^2$. Since the square of a real number is nonnegative, the only way this equation could be true is if both sides are $0$: $a=c$ and $b=d$.
This shows that there are no redundancies, and every number of the form $a+bi$ is distinct.
Addition and subtraction
The next thing to worry about is whether we need more numbers to be able to add and subtract. Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$, and that if $a,b,c,d$ are real, then so are $a\pm c$ and $b\pm d$. So we don't get a new type of number when we add or subtract.
Vectors?
The above calculation says that when you add or subtract complex numbers written in the $a+bi$ form, you add/subtract the $a$'s and you add and subtract the $b$'s. This is just like the analogous operations on vectors: $\langle a,b\rangle\pm\langle c,d\rangle=\langle a\pm c,b\pm d\rangle$.
Multiplication
What about multiplication? Note that $(a+bi)(c+di)=ac+bci+adi+bdi^2=(ac-bd)+(ad+bc)i$ since $i^2=-1$. So when we multiply two numbers of this form, we get back a number that can be written in this form (even though it can also be written in other forms like one with $i^2$ in it).
Vectors?
If we look at the above calculation in the special case when $d=0$, it says $c(a+bi)=(ac)+(bc)i$. So when you multiply complex numbers written in $a+bi$ form by a real number, you just multiply the $a$ and the $b$ separately. This is just like the analogous scalar multiplication operation on 2-dimensional real vectors: $c\langle a,b\rangle=\langle ca,cb\rangle$.
So as far as addition, subtraction, and multiplication by real numbers is concerned, complex numbers are just like 2-d real vectors.
Division
What about division?
$$\begin{align*}&\frac{a+bi}{c+di} \\ &= \frac{(a+bi)(c-di)}{(c+di)(c-di)} \\ &= \frac{ac+bci-adi-bdi^2}{c^2-(di)^2} \\ &= \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ &= \left(\frac{ac+bd}{c^2+d^2}\right)+\left(\frac{bc-ad}{c^2+d^2}\right)i\end{align*}$$
This looks like it works, but we should worry about whether the denominator is zero. Since $c$ and $d$ are real numbers, their squares are nonnegative, so $c^2+d^2$ could only be zero if we had $c=d=0$. But that would make the original denominator $c+di=0+0i=0$. We don't expect division by zero to work, so division of these numbers works fine whenever it should.
