Math Assignment Help With Multiplication properties
1.6.2 Multiplication properties:
Theorem1: (Closure property): If z1 and z2 ϵ C, then the product of two complex numbers is a complex number, i.e.
z1. z2 ϵ C
Proof: Let z1 = (a + ib)
And,
z2 = c + id
where
a, b, c, d ϵ R
therefore,
z1 . z2 = (a + ib) (c + id)
= (ac – bd) + i (ad + bc) ϵ C
As stated by the closure property of addition, subtraction and multiplication of real numbers (ac – bd) and (ad + bc) both are real numbers.
Hence,
(ac – bd) + i (ad + bc) ϵ C
Therefore,
z1 ϵ C ; z2 ϵ C z1. z2 ϵ C for all z1, z2 ϵ C
Theorem 2 :( Commutative Law): If z1, z2 ϵ C then,
z1z2 = z2z1
Proof: Let z1 = (a + ib)
And,
z2 = c + id
z1 z2 = (a + ib) (c + id)
= (ac – bd) + i (ad + bc)
= (ca – db) + i (da + cb)
= (c + id) (a + ib)
= z2 z1
Thus,
z1z2 = z2z1
Theorem3: (Associative law): If z1, z2, z3 ϵ C, then
(z1 z2) z3 = z1 (z2 z3)
Proof: Let z1 = a + ib
And,
z2 = c + id
z3 = e + if
(z1 z2) z3 = [(a + ib) (c + id)] (e + if)
= [(ac – bd) + i (ad + bc)] + (e +if)
= {(ac – bd) e – (ad + bc) f}+ i {(ac – bd)f + (ad +bc)e}
= (ace – bde – adf – bcf) + i (acf – bdf + ade + bce)
= {a (ce – df) – b (cf + de)} + i {a(cf + de) + b(ce – df)}
= (a + ib) {(ce – df) +i (cf +de)}
= z1 (z2 z3)
Thus,
(z1 z2) z3 = z1 (z2 z3)
Theorem4: (Multiplicative Inverse): For each z ϵ C, there exist a complex number denoted by z-1 such that
z z-1 = z-1 z = 0
Proof: z = a + ib ϵ C
And,
z ≠ 0
let z-1 = (x + iy) ϵ C
Then,
z z-1 = 1
(a + ib) (x + iy) = 1
(ax – by) + i(bx + ay) = 1 + i0
ax – by = 1
and
bx + ay = 0
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