## Theorem 1.1.8: Complex Numbers are a Field |

The set of complex numbers |

We need to prove the field axioms for our definition of addition and multiplication:

*z + w = (x,y) + (u,v) = (x+u, y+v)**z * w = (x,y) * (u,v) = (x*u - y*v, x*v + y*u)*

1. Both *+* and *** are associative, which is obvious
for addition. For multiplication we nned to show that *a*(b*c)=(a*b)*c*.
Let
*a=(x _{1},y_{1})*,

2. Both *+* and *** are commutative, i.e.
*a+b=b+a* and *a*b=b*a*

Exercise

3. The distributive law holds, i.e. *a*(b+c)=(a*b)+(a*c)*

Exercise

4. The additive identity is *(0,0)*, and the multiplicative
identity is *(1,0)*, which you can easily confirm.

5. The additive inverse to *(x,y)* is *(-x,-y)*. The
multiplicative inverse to *(x,y)* is
*(x/(x ^{2}+y^{2}), -y/(x^{2}+y^{2})*.

That shows that * C* is a field.

We now identify the real number *x* with the complex number
*(x,0)*. Thus, the (real) numbers 0 and 1 are the same as the
complex numbers *(0,0)* and *(1,0)*, respectively.
Technically, the map

defined viaf:{(x,y)R: y = 0}C

The complex number *(0,1)* has the property that
*(0,1)*(0,1)=(-1,0)*, which is the same property as our symbol
*i*.