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(Word problem)How can the associative , commutative , and distributive properties be applied when performing operations on complex numbers and polynomials ?

User Newcoma
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1. Complex numbers:

Let's probe associative, commutative and distributive properties for any three complex numbers:

Let Z1=a+ix, Z2=b+iy and Z3=c+iz

a. Associative property: (z1+z2)+z3=z1+(z2+z3)

Then:

(z1+z2)+z3=((a+ix)+(b+iy))+c+iz

=(a+b+ix+iy)+c+iz

=(a+b+i(x+y))+c+iz

=a+b+c+i(x+y+z))

=a+b+c+ix+iy+iz

=a+ix+(b+c+iy+iz)

=a+ix+((b+iy)+(c+iz))

=z1+(z2+z3)

Hence, complex numbers follow the associative property.

b. Commutative property: z1+z2=z2+z1

Then:

z1+z2=(a+ix)+(b+iy)

=a+ix+b+iy

=b+iy+a+ix

=(b+iy)+(a+ix)

=z2+z1

Hence, complex numbers follow the commutative property.

c. Distributive property: z1(z2+z3)=z1*z2+z1*z3

z1*(z2+z3)=(a+ix)[(b+iy)+(c+iz)]

=(a+ix)*(b+iy)+(a+ix)(c+iz)

=z1*z2+z1*z3

Hence, complex numbers follow the distributive property.

2. Polynomials:

Let P=ax+b, Q=ax^2+bx+c, R=bx+c

a. Associative property: (P+Q)+R=P+(Q+R)

(P+Q)+R=[(ax+b)+(ax^2+bx+c)]+bx+c

=[ax^2+ax+bx+b+c]+bx+c

=(ax+b)+[(ax^2+bx+c)+(bx+c)]

=P+(Q+R)

Polynomials follow the associative property

b. Commutative property: P+R=R+P

P+R=(ax+b)+(bx+c)

=ax+b+bx+c

=bx+c+ax+b

=(bx+c)+ax+b)

=R+P

Polynomials follow the commutative property.

d. Distributive property: P(Q+R)=PQ+QR

P(Q+R)=(ax+b)[(ax^2+bx+c)+(bx+c)]

=ax^3+abx^2+acx+abx^2+acx+abx^2+b^2x+bc+b^2x+bc

=ax^3+3abx^2+2acx+2b^2x+2bc

PQ+QR=(ax+b)(ax^2+bx+c)+(ax+b)(bx+c)

=ax^3+abx^2+acx+abx^2+b^2x+bc+abx^2+acx+b^2x+bc

=ax^3+3abx^2+2acx+2b^2x+2bc

Then:

P(Q+R)=PQ+PR

Polynomials follow the distributive property.

User Retif
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