Properties of Complex Numbers – Part 1

Just like real numbers, complex numbers follow certain algebraic rules. In this part, we'll investigate whether complex numbers satisfy the foundational properties of arithmetic: commutativity, associativity, and distributivity. These are the building blocks of algebra, and seeing them in the context of complex numbers helps us better understand their structure and reliability. Part 2 will take this further by exploring the identities and inverses of complex numbers.

Commutative Property

Let, z1=a+bi z_1 = a+bi and z2=c+diz_2 = c+di

For Addition

In mathematics, the commutative property tells us that the order in which we add numbers doesn't affect the result. But does this rule extend to complex numbers? Let's explore it with a direct and logical derivation.

Prove That z1+z2=z2+z1z_1 + z_2 = z_2 + z_1 (Commutative Property for Addition)

z1+z2=z2+z1z_1 + z_2 = z_2 + z_1

LHS=z1+z2LHS= z_1 + z_2

=(a+bi)+(c+di)= (a + bi) + (c + di)

=a+bi+c+di= a + bi + c + di

=a+c+bi+di[Commutative Property of R]= a + c + bi + di \quad \text{[Commutative Property of } \mathbb{R}\text{]}

=(a+c)+(b+d)i(1)= (a + c) + (b + d)i \quad \quad (1)

z1+z2=z2+z1z_1 + z_2 = z_2 + z_1

RHS=z2+z1RHS= z_2 + z_1

=(c+di)+(a+bi)= (c + di) + (a + bi)

=c+di+a+bi= c + di + a + bi

=c+a+di+bi[Commutative Property of R]= c + a + di + bi \quad \text{[Commutative Property of } \mathbb{R}\text{]}

=(c+a)+(d+b)i= (c + a) + (d + b)i

=(a+c)+(b+d)i(2)[Commutative Property of R]= (a + c) + (b + d)i \quad \quad (2) \quad \text{[Commutative Property of } \mathbb{R}\text{]}

SKIP

From (1) and (2)

⇒LHS=RHS\Rightarrow LHS = RHS

For Multiplication

Now let's consider multiplication. Does the order of multiplying two complex numbers change the outcome? As with real numbers, the answer is no – and here's the step-by-step demonstration that confirms this.

Prove That z1⋅z2=z2⋅z1z_1 \cdot z_2 = z_2 \cdot z_1 (Commutative Property for Multiplication)

z1⋅z2=z2⋅z1z_1 \cdot z_2 = z_2 \cdot z_1

LHS=z1⋅z2LHS = z_1 \cdot z_2

(a+bi)(c+di)(a + bi)(c + di)

=a(c+di)+bi(c+di)[Distributive Property for R]= a(c + di) + bi(c + di) \quad \text{[Distributive Property for } \mathbb{R}\text{]}

=ac+adi+bci+bdi2[Distributive Property for R]= ac + adi + bci + bdi^2 \quad \text{[Distributive Property for } \mathbb{R}\text{]}

=ac+bdi2+adi+bci[Commutative Property for R]= ac + bd i^2 + adi + bci \quad \text{[Commutative Property for } \mathbb{R}\text{]}

=ac+bd⋅(−1)+i(ad+bc)where i2=−1= ac + bd \cdot (-1) + i(ad + bc) \quad \text{where } i^2 = -1

=(ac−bd)+(ad+bc)i(1)= (ac - bd) + (ad + bc)i \quad \quad (1)

SKIP

Now let's evaluate the right-hand side.

RHS=z2⋅z1RHS = z_2 \cdot z_1

(c+di)(a+bi)(c + di)(a + bi)

=c(a+bi)+di(a+bi)[Distributive Property for R]= c(a + bi) + di(a + bi) \quad \text{[Distributive Property for } \mathbb{R}\text{]}

=ca+cbi+dai+dbi2[Distributive Property for R]= ca + cbi + dai + dbi^2 \quad \text{[Distributive Property for } \mathbb{R}\text{]}

=ca+dbi2+dai+cbi[Commutative Property for R]= ca + db i^2 + dai + cbi \quad \text{[Commutative Property for } \mathbb{R}\text{]}

=ca+db⋅(−1)+i(da+cb)where i2=−1= ca + db \cdot (-1) + i(da + cb) \quad \text{where } i^2 = -1

=(ca−db)+(da+cb)i= (ca - db) + (da + cb)i

=(ac−bd)+(ad+bc)i(2)[Commutative Property for R]= (ac - bd) + (ad + bc)i \quad \quad (2) \quad \text{[Commutative Property for } \mathbb{R}\text{]}

SKIP

From (1) and (2)

⇒LHS=RHS\Rightarrow LHS = RHS

Associative Property

Let, z1=a+bi z_1 = a+bi, z2=c+diz_2 = c+di, and z3=e+fiz_3 = e+fi

For Addition

The associative property of addition reassures us that how we group terms when adding three or more numbers doesn't change the final result. Does this hold true when those numbers are complex? Absolutely. Let's confirm this through an elegant derivation.

Prove That (z1+z2)+z3=z1+(z2+z3)(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) (Associative Property for Addition)

(z1+z2)+z3=z1+(z2+z3)(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)

LHS=(z1+z2)+z3LHS = (z_1 + z_2) + z_3

=((a+bi)+(c+di))+(e+fi)= ((a + bi) + (c + di)) + (e + fi)

=(a+c+bi+di)+(e+fi)= (a + c + bi + di) + (e + fi)

=(a+c+(b+d)i)+(e+fi)= (a + c + (b + d)i) + (e + fi)

=a+c+(b+d)i+e+fi= a + c + (b + d)i + e + fi

=a+c+e+(b+d)i+fi= a + c + e + (b + d)i + fi

=(a+c+e)+(b+d+f)i(1)= (a + c + e) + (b + d + f)i \quad \quad (1)

SKIP

Now we evaluate the right-hand side.

RHS=z1+(z2+z3)RHS = z_1 + (z_2 + z_3)

=(a+bi)+((c+di)+(e+fi))= (a + bi) + ((c + di) + (e + fi))

=(a+bi)+(c+e+di+fi)= (a + bi) + (c + e + di + fi)

=(a+bi)+(c+e+(d+f)i)= (a + bi) + (c + e + (d + f)i)

=a+bi+c+e+(d+f)i= a + bi + c + e + (d + f)i

=a+c+e+bi+(d+f)i= a + c + e + bi + (d + f)i

=(a+c+e)+(b+d+f)i(2)= (a + c + e) + (b + d + f)i \quad \quad (2)

SKIP

From (1) and (2)

⇒LHS=RHS\Rightarrow LHS = RHS

For Multiplication

When multiplying several numbers, does it matter how we group them? For complex numbers, the answer is again reassuring: no, it does not. This is the associative property of multiplication at work — and here's the formal proof.

Prove That (z1⋅z2)⋅z3=z1⋅(z2⋅z3)(z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3) (Associative Property for Multiplication)

(z1⋅z2)⋅z3=z1⋅(z2⋅z3)(z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3)

LHS=(z1⋅z2)⋅z3LHS = (z_1 \cdot z_2) \cdot z_3

=((a+bi)(c+di))⋅(e+fi)= ((a + bi)(c + di)) \cdot (e + fi)

=((ac−bd)+(ad+bc)i)⋅(e+fi)= ((ac - bd) + (ad + bc)i) \cdot (e + fi)

=[(ac−bd)e−(ad+bc)f]+[(ac−bd)f+(ad+bc)e]i= [(ac - bd)e - (ad + bc)f] + [(ac - bd)f + (ad + bc)e]i

=[ace−bde−adf−bcf]+[acf−bdf+ade+bce]i(1)= [ace -bde - adf - bcf] + [acf - bdf + ade + bce]i \quad \quad (1)

SKIP

Now we evaluate the right-hand side.

RHS=z1⋅(z2⋅z3)RHS = z_1 \cdot (z_2 \cdot z_3)

=(a+bi)⋅((c+di)(e+fi))= (a + bi) \cdot ((c + di)(e + fi))

=(a+bi)⋅((ce−df)+(cf+de)i)= (a + bi) \cdot ((ce - df) + (cf + de)i)

=[(a)(ce−df)−(b)(cf+de)]+[(a)(cf+de)+(b)(ce−df)]i= [(a)(ce - df) - (b)(cf + de)] + [(a)(cf + de) + (b)(ce - df)]i

=[ace−adf−bcf−bde]+[acf+ade+bce−bdf]i= [ace - adf - bcf - bde] + [acf + ade + bce - bdf]i

=[ace−bde−adf−bcf]+[acf−bdf+ade+bce]i(2)= [ace -bde - adf - bcf] + [acf - bdf + ade + bce]i \quad \quad (2)

SKIP

From (1) and (2)

⇒LHS=RHS\Rightarrow LHS = RHS

Distributive Property

The distributive property links addition and multiplication: multiplying a number across a sum gives the same result as multiplying each term individually and then adding the results. This principle is essential in algebra, and it’s comforting to know that complex numbers also obey this property. Let’s walk through the logic to see it in action.

Prove That z1(z2+z3)=z1z2+z1z3z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3 (Distributive Property)

z1(z2+z3)=z1z2+z1z3z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3

LHS=z1(z2+z3)LHS = z_1(z_2 + z_3)

=(a+bi)[(c+di)+(e+fi)]= (a + bi)[(c + di) + (e + fi)]

=(a+bi)[(c+e)+(d+f)i][Grouping like terms]= (a + bi)[(c + e) + (d + f)i] \quad \text{[Grouping like terms]}

=a[(c+e)+(d+f)i]+bi[(c+e)+(d+f)i][Distributive Property]= a[(c + e) + (d + f)i] + bi[(c + e) + (d + f)i] \quad \text{[Distributive Property]}

=a(c+e)+a(d+f)i+bi(c+e)+bi(d+f)i[Distributive Property]= a(c + e) + a(d + f)i + bi(c + e) + bi(d + f)i \quad \text{[Distributive Property]}

=ac+ae+adi+afi+bci+bei+bdi2+bfi2= ac + ae + adi + afi + bci + bei + bdi^2 + bfi^2

=ac+ae+adi+afi+bci+bei−bd−bf= ac + ae + adi + afi + bci + bei - bd - bf

=(ac+ae−bd−bf)+(ad+af+bc+be)i(1)= (ac + ae - bd - bf) + (ad + af + bc + be)i \quad \quad (1)

SKIP

Now let's evaluate the RHS

RHS=z1z2+z1z3RHS = z_1 z_2 + z_1 z_3

=(a+bi)(c+di)+(a+bi)(e+fi)= (a + bi)(c + di) + (a + bi)(e + fi)

=ac+adi+bci+bdi2+ae+afi+bei+bfi2= ac + adi + bci + bdi^2 + ae + afi + bei + bfi^2

=ac+adi+bci−bd+ae+afi+be−bf[Because i2=−1]= ac + adi + bci -bd + ae + afi + be - bf \quad \text{[Because } i^2 = -1]

=(ac+ae−bd−bf)+(ad+af+bc+be)i(2)= (ac + ae - bd - bf) + (ad + af + bc + be)i \quad \quad (2)

SKIP

From (1) and (2)

⇒LHS=RHS\Rightarrow LHS = RHS


End of Lesson

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Multiplication of Complex Numbers
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Properties of Complex Numbers II