Math Assignment Help With Complex Number
1.6 Algebra of Complex Numbers:
Complex numbers are added, subtracted, and multiplied in a similar way. To simplify complex-valued expressions, combine like terms and apply the various other methods that we have learned for working with polynomials.
1.6.1 Addition Properties:
Theorem1 :( Closure property): If z1 and z2 ϵ C, then the sum of two complex numbers is a complex number, i.e.
z1 + z2 ϵ C
Proof: let z1, z2 ϵ C
Where, z1 = a +ib
And,
z2 = c + id
where
a, b, c, d ϵ R
therefore,
z1 + z2 = (a + ib) + (c + id)
= (a + c) + i (b + d) ϵ C
Hence,
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,
z1 + z2 = z2 + z1
Proof: let z1 = a + ib,
z2 = c + id
then,
z1 + z2 = (a + ib) + (c + id)
= (a + c) + i (b + d)
= (c + a) + i (d + b)
= (c + id) + (a + ib)
= z2 + z1
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)
= [(a + c) + i (b + d)] + (e +if)
= [(a + c + e) + i (b + d + f)]
{`= {a + (c + e)} + i {b + (d + f)}`}
{`= (a + ib) + {(c +e) +i (d + f)}`}
{`= z1 + (z2 + z3)`}
Theorem4: (Additive inverse): For each z ϵ C, there exist –z ϵ C such that
z + (- z) = (- z) + z = 0
Proof: z = a + ib
Then,
-z = (-a) + i (-b) ϵ C
Therefore,
z + (-z) = (a + ib) + [(-a) + i (-b)]
= [a + (-a)] + i [b + (-b)]
= 0 + i 0
= 0
Thus,
z + (- z) = 0
Similarly,
(-z) + z = 0
Therefore,
z + (-z) = (-z) + z = 0
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