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.

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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