4 1. The Complex Numbers

0

1 3 2 4

i

2i

.

3+2i

−1

−i

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

R2,

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),

while

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

and

(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,