Definitions for complex numbers I could not find this question anywhere else. But why are addition, subtraction, division, and other operations defines they are in complex numbers? Could they defined as something else?
 A: To add $a$ and $b$, start at $0$ and draw an arrow to $a$ and another arrow to $b$.  Then move the arrow from $0$ to $b$ so that its starting point is at $a$.  Then its ending point is at $a+b$.
That works BOTH for real numbers and for complex numbers.  So you're doing the same thing as with real numbers, but in a plane rather than in a line.
To multiply $a$ and $b$, draw the line through $0$ and $a$, and leave $0$ where it is but write $1$ where $a$ is.  Having put $0$ and $1$ in their places locate $b$ in this stretched coordinate system.  Look at what number appears at that point in the original coordinate system before stretching.  That number is $a\times b$.
For example, say $a=3$ and $b=2$.  Put $1$ where $a=3$ is, then see where $b=2$ is in the stretched system.  It's at $6$.  Therefore $a\times b=3\times 2=6$:
$$
\begin{array}{ccccccccccccccccccccccc}
0 & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & 7 & \cdots \\
| & \qquad & | & \qquad & | & \qquad & | & \qquad & | & \qquad & | & \qquad & | & \qquad & | & \cdots \\
0 & & & & & & 1 & & & & & & 2 & & & \cdots
\end{array}
$$
That way of multiplying works BOTH in the real numbers and in the complex numbers.  So you're doing the same thing as with real numbers, but in a plane rather than in a line.  Anyone who's ever wondered why the product of two negative numbers is positive can use this idea to see the answer.  This also explains why multiplication by $i$ amounts to a $90^\circ$ counterclockwise rotation.
A: Yes, they could, at least for multiplication.  You can define $(x,y)\cdot(z,w)$ to be $(xz-3yw,xw+yz)$ for every $x,y,z,w\in\mathbb{R}$.  How deep is your knowledge about fields and field extensions?

EDIT:
Basically, if you want a field extension of $\mathbb{R}$, then this field will be a vector space over $\mathbb{R}$, so you don't have liberty with defining addition on this field extension as it has to be carried over from the addition on $\mathbb{R}$.  If the extension is algebraic and proper, then you can construct it by consider the quotient $\mathbb{R}[t]/\big(f(t)\big)$ for some polynomial $f(t)\in\mathbb{R}[t]$ irreducible over $\mathbb{R}$.  Such polynomials are of the form $f(t)=at^2+bt+c$ with $a,b,c\in\mathbb{R}$ are such that $a\neq 0$ and $b^2-4ac<0$.  If you define $\bar{t}$ to be $t+\big(f(t)\big)$ in $\mathbb{R}[t]/\big(f(t)\big)$, then every element of $\mathbb{R}[t]/\big(f(t)\big)$ can be written as $u+v\bar{t}$ for some $u,v\in\mathbb{R}$.  Note that $(x+y\bar{t})(z+w\bar{t})=\left(xz-\frac{c}{a}yw\right)+\left(xw+yz-\frac{b}{a}yw\right)\bar{t}$ is the multiplication on $\mathbb{R}[t]/\big(f(t)\big)$ (this is due to the fact that $a\bar{t}^2+b\bar{t}+c=0$ in $\mathbb{R}[t]/\big(f(t)\big)$).
The usual choice of this polynomial is $f(t)=t^2+1$, then you have the usual operations of the complex numbers.  In my example, I chose $f(t)=t^2+3$, but you can choose many other different polynomials such as $f(t)=t^2+t+41$, where your multiplication would become $(x+y\bar{t})(z+w\bar{t})=(xz-41yw)+(xw+yz-yw)\bar{t}$.  However, no matter which polynomial $f(t)$ you take, you will end up with a field isomorphic to $\mathbb{C}$.  A field isomorphism $\varphi:\mathbb{R}[t]/\big(f(t)\big)\to\mathbb{C}$ is given by $u+v\bar{t}\mapsto\left(u-\frac{b}{2a}v\right)+\left(\frac{\sqrt{4ac-b^2}}{2a}v\right)\text{i}$.  It is just that $\mathbb{C}=\mathbb{R}[t]/\big(t^2+1\big)$ is the easiest algebraic extension of $\mathbb{R}$ to deal with.
A: One way to view complex numbers is that they are an attempt to make $\mathbb{R}^2$ into a field while still retaining some of the fundamental properties of $\mathbb{R}$. As such, we would want to keep vector addition the same when translating between $\mathbb{R}^2$ and $\mathbb{C}$. Hence $(w,x)+(y,z) = (w+y,x+z)$. From there you can get subtraction quite easily since subtraction is just adding the additive inverse.
The tricky part is the multiplication. You can always define a multiplication on vector spaces but it might not really be a "useful" construct. For instance one way we could multiply vector elements is by $(w,x)\cdot(y,z) = (wy,xz)$, i.e. componentwise multiplication. The problem with this is that this doesn't give $\mathbb{R}^2$ a field structure. Particularly, there are nonzero elements which multiply to $(0,0)$: $(1,0)$ and $(0,1)$.
So the simplest construction fails pretty spectacularly. The next simplest might be $(w,x)\cdot(y,z) = (wz,xy)$. However again there are nonzero elements which multiply to $(0,0)$: $(1,0)$ and $(1,0)$. So even this doesn't work.

This is the point where we turn to the properties of $\mathbb{R}$ and make one "modest" geometric assumption.


*

*We can identify $\mathbb{R}$ with the $x$ axis in $\mathbb{R}^2$. If we were to respect the field properties of $\mathbb{R}$, then we would need that $(x,0)\cdot(y,0) = (xy,0)$ - meaning if we multiply two complex numbers with $0$ in the second component (i.e. two real numbers), we should get a complex number with $0$ in its second argument (i.e. a real number).

*Another piece is noting that $\mathbb{R}^2$ is a vector space over $\mathbb{R}$. Meaning that if we have real $\alpha$, then $\alpha(x,y) = (\alpha x,\alpha y)$. In terms of complex numbers, $(\alpha, 0)\cdot(x,y) = (\alpha x,\alpha y)$.

*Since it doesn't matter in which order we multiply real numbers (i.e. $xy = yx$), it seems reasonable to want this for complex numbers so we will impose this condition as well.

*If $(w,x)$ is a unit vector, then $(w,x)\cdot(y,z)$ has the same length as $(y,z)$. 
The first three points heavily constrain the sort of multiplication we can do between complex numbers. If we multiplied $(w,x)$ and $(y,z)$, what should emerge is something of the form$^{1}$
$$\left(\begin{array}{c}\alpha_1 wy + \alpha_2 wz + \alpha_3 xy + \alpha_4 xz \\ \beta_1 wy + \beta_2 wz + \beta_3 xy + \beta_4 xz\end{array}\right).$$
If $x = 0 = z$, then we fall into the case of condition $1$ above and our multiplication should result in $(wy, 0)$, meaning that $\alpha_1 = 1$ and $\beta_1 = 0$.
If $x = 0$, then we fall into the case of condition $2$ above. In which case the product should become $(wy, wz)$. This says that $\alpha_2 = 0 = \beta_1$ and $\beta_2 = 1$.
This simplifies our multiplication a bit:
$$\left(\begin{array}{c} w \\ x \end{array}\right)\cdot\left(\begin{array}{c} y \\ z\end{array}\right) = \left(\begin{array}{c} wy + \alpha_3 xy + \alpha_4 xz \\ wz + \beta_3 xy + \beta_4 xz\end{array}\right).$$
Making use of commutativity (condition $3$),
$$\left(\begin{array}{c} w \\ x\end{array}\right)\cdot\left(\begin{array}{c} y \\ z\end{array}\right) = \left(\begin{array}{c} y \\ z\end{array}\right)\cdot \left(\begin{array}{c} w \\ x\end{array}\right)$$
or equivalently (using our simplified multiplication)
$$\left(\begin{array}{c} wy + \alpha_3 xy + \alpha_4 xz \\ wz + \beta_3 xy + \beta_4 xz\end{array}\right) = \left(\begin{array}{c} wy + \alpha_3 wz + \alpha_4 xz \\ xy + \beta_3 wz + \beta_4 xz\end{array}\right)$$
On the left hand side, there is no appearance of $wz$ in the first component so $\alpha_3 = 0$. Moreover, $\beta_3 = 1$ to match the $\beta_3 wz$ in the second component on the right hand side and the $wz$ in the second component on the left hand side. This simplifies our multiplication to
$$\left(\begin{array}{c} w \\ x\end{array}\right)\cdot\left(\begin{array}{c} y \\ z\end{array}\right) = \left(\begin{array}{c} wy + \alpha_4 xz \\ xy + wz + \beta_4 xz\end{array}\right).$$

So now we have two parameters which we need to somehow do away with. The reason for these parameters existing in the first place is that we have really only used the real structure of complex numbers; we haven't given any meaning to the multiplication of two pure imaginary numbers (vectors of the form $(0,a)$), hence the appearance of $xz$ in both the numerator and denominator.
At this point, we have invoked most of the properties of a field but we are still missing one: no zero divisors, meaning no two nonzero elements multiply to zero.
Taking $(w,x)$ to be a fixed nonzero vector, we can view $(w,x)\cdot(y,z)$ as a matrix product:
$$\left(\begin{array}{cc} w & \alpha_4 x \\ x & w+\beta_4 x\end{array}\right)\left(\begin{array}{c} y \\ z\end{array}\right).$$
If we don't want $(w,x)\cdot(y,z)$ to be zero, then the nullspace of the above matrix must be trivial. That is to say that the determinant of the above matrix must be nonzero. Equivalently, if the determinant of the above matrix is zero, then both $w$ and $x$ are zero.
Taking the determinant and setting it to zero, we get $w^2 + \beta_4 wx - \alpha_4x^2 = 0$. Note that both $\alpha_4$ and $\beta_4$ cannot be zero since then $(0,x)\cdot(0,z) = (0,0)$ for arbitrary $x,z$. If $x=0$, then $w$ must be $0$. Consider then $x\neq 0$.
The only way for there to be no solution to the above equation (in real numbers which is what $w$ and $x$ are) is if the discriminant is less than zero, i.e. $\beta_4^2x^2 + 4\alpha_4x^2 < 0$. Since $x^2 > 0$, this simplifies to $\beta_4^2 + 4\alpha_4 < 0$. Thus any $\alpha_4$ satisfying $\alpha_4 < -\left(\frac{\beta_4}{2}\right)^2$ will give rise to a field with $\beta_4$ being a free parameter.

This is a little bit unsatisfactory since we can't "derive" the complex number structure just from our first three conditions above. These conditions do not fully encompass the spirit of complex numbers. Complex numbers are meant to be not only algebraic in their origin but also geometric. This is the reason for our fourth assumption - to capture the geometric flavor of complex numbers.
If we invoke our geometric condition (condition $4$), we can completely solve the problem. Taking $w= 0 = y$ and $x=1 = z$, then $(0,1)\cdot(0,1) = (\alpha_4 ,\beta_4 )$. Thus $\alpha_4^2 + \beta_4^2 = 1$.
If $(w,x)$ is a unit vector, then $w = \cos\theta$ and $y = \sin\theta$ for some angle $\theta$. If $(w,x)\cdot(y,z)$ is to have the same length as $(y,z)$, then the matrix $(w,x)$ must have determinant $1$, meaning that
$$w^2 + \beta_4wx -\alpha_4 x^2 = 1 \Longrightarrow \cos^2\theta + \beta_4 \sin\theta\cos\theta -\alpha_4\sin^2\theta = 1.$$
Substituting $\alpha_4 = \pm\sqrt{1-\beta_4^2}$, this becomes
$$\cos^2\theta + \beta_4\sin\theta\cos\theta \mp \sqrt{1-\beta_4^2} \sin^2\theta = 1$$
which further reduces to
$$\left(\sqrt{1-\beta_4^2}-1\right)\sin^2\theta + \beta_4\sin\theta\cos\theta = 0.$$
Picking $\theta = \frac{\pi}{4}$, we see that $\sin\theta\cos\theta > 0$ and so $\beta_4 = 0$.
Thus $\alpha_4 = \pm 1$ but the only way for $(w,x)\cdot(y,z)$ to have the same length as $(y,z)$ (when $(w,x)$ is a unit vector) is if $\alpha_4 = -1$. Hence the product of $(w,x)$ and $(y,z)$ is $(wy - xz, wz+xy)$ which is exactly the algebraic structure of the complex numbers.
From this, you of course get that $(0,1)\cdot(0,1) = (-1,0)$ for free, which is of course the identity $i^2 = -1$. You also get to see first hand how the matrix form of complex numbers works (since it's more or less built into my "derivation") and by proxy the polar form and what the othe field structures on $\mathbb{R}^2$ are.

$^{1}$ You could of course have higher powers but then you'd lose linearity of multiplication over addition.
A: We define in $\Bbb R^2$ the operations $+$ and $\cdot$ by: $$(a,b)+(c,d) = (a+c,b+d), \quad (a,b)\cdot(c,d) = (ac-bd, ad+bc)$$because these operations make $\Bbb C \stackrel{\rm def}{=} (\Bbb R^2, +, \cdot)$ a field. One can prove that: $$\Bbb R \ni x \hookrightarrow (x,0) \in \Bbb R^2$$is an isomorphism onto its image, so that $\Bbb C$ contains a "copy" of $\Bbb R$ inside it. Also, note that $(0,1)\cdot (0,1) = (-1,0)$. From this, we call $(0,1) \equiv i$. Then the notation: $$(x,y) = (x,0)+(0,y)= (x,0) + (y,0)(0,1) \equiv x+iy.$$You could define other operations on $\Bbb R^2$, but most likely they won't have nice properties as the ones above.
A: The definition of arithmetic of complex numbers is forced on you if you want it to extend the ordinary arithmetic of real numbers and still satisfy the usual "laws of arithmetic" (things like $x+y=y+x$, $x(y+z)=xy+xz$, etc.).  For instance, suppose you have complex numbers $z=a+bi$ and $w=c+di$, where $a,b,c,d\in\mathbb{R}$.  Then if we assume addition and multiplication satisfies all the usual rules, we get
$$z+w=(a+bi)+(c+di)=(a+c)+(bi+di)=(a+c)+(b+d)i$$
and
$$zw=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i.$$
By similar manipulations, you can obtain the formulas for subtraction and division of complex numbers (though the division one is trickier to derive).
More precisely, the "usual rules" here are that addition and multiplication satisfy the axioms of a commutative ring (and subtraction and division are defined using additive and multiplicative inverses).  The above derivations can be interpreted as a proof that in any commutative ring $C$ containing the real numbers as a subring and an element $i\in C$ such that $i^2=-1$, addition and multiplication of elements of the form $a+bi$ for $a,b\in\mathbb{R}$ must satisfy the formulas above.
