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,
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