4 1. The Complex Numbers
1 3 2 4
Figure 1.1.1. Plot of the Complex Number 3 + 2i.
(e) z1(z2 + z3) = z1z2 + z1z3 – distributive law.
Proof. As far as the operation of addition is concerned, C is just
which is a
vector space over R. Parts (a) and (b) of the theorem follow directly from this.
Part (c) is obvious from the definition of multiplication. We will prove part (d) and
leave part (e) as an exercise (Exercise 1.1.8).
If zj = xj + iyj for j= 1, 2, 3, then
(z1z2)z3 = ((x1 + iy1)(x2 + iy2))(x3 + iy3)
= (x1x2 − y1y2 + i(x1y2 + y1x2))(x3 + iy3)
= x1x2x3 − y1y2x3 − x1y2y3 − y1x2y3
+ i(x1x2y3 − y1y2y3 + x1y2x3 + y1x2x3),
z1(z2z3) = (x1 + iy1)((x2 + iy2)(x3 + iy3))
= (x1 + iy1)(x2x3 − y2y3 + i(x2y3 + y2x3))
= x1x2x3 − x1y2y3 − y1x2y3 − y1y2x3
+ i(x1x2y3 + x1y2x3 + y1x2x3 − y1y2y3).
Since the results are the same, the proof of (d) is complete.
The properties described in the above theorem are some of the properties that
must hold in a field. A field must also have additive and multiplicative identities –
that is, elements 0 and 1 which satisfy
(1.1.3) z + 0 = z
(1.1.4) 1 · z = z
for every element z in the field. That this holds for C follows immediately from the
fact that C is a vector space over R.
A field must also have the properties that every element z has an additive
inverse, that is, an element −z such that
(1.1.5) z + (−z) = 0,