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 and z2=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+z1 (Commutative Property for Addition)
z1+z2=z2+z1
LHS=z1+z2
=(a+bi)+(c+di)
=a+bi+c+di
=a+c+bi+di[Commutative Property of R]
=(a+c)+(b+d)i(1)
z1+z2=z2+z1
RHS=z2+z1
=(c+di)+(a+bi)
=c+di+a+bi
=c+a+di+bi[Commutative Property of R]
=(c+a)+(d+b)i
=(a+c)+(b+d)i(2)[Commutative Property of R]
SKIP
From (1) and (2)
⇒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⋅z1 (Commutative Property for Multiplication)
z1⋅z2=z2⋅z1
LHS=z1⋅z2
(a+bi)(c+di)
=a(c+di)+bi(c+di)[Distributive Property for R]
=ac+adi+bci+bdi2[Distributive Property for R]
=ac+bdi2+adi+bci[Commutative Property for R]
=ac+bd⋅(−1)+i(ad+bc)where i2=−1
=(ac−bd)+(ad+bc)i(1)
SKIP
Now let's evaluate the right-hand side.
RHS=z2⋅z1
(c+di)(a+bi)
=c(a+bi)+di(a+bi)[Distributive Property for R]
=ca+cbi+dai+dbi2[Distributive Property for R]
=ca+dbi2+dai+cbi[Commutative Property for R]
=ca+db⋅(−1)+i(da+cb)where i2=−1
=(ca−db)+(da+cb)i
=(ac−bd)+(ad+bc)i(2)[Commutative Property for R]
SKIP
From (1) and (2)
⇒LHS=RHS
Associative Property
Let, z1=a+bi, z2=c+di, and z3=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) (Associative Property for Addition)
(z1+z2)+z3=z1+(z2+z3)
LHS=(z1+z2)+z3
=((a+bi)+(c+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+e+(b+d)i+fi
=(a+c+e)+(b+d+f)i(1)
SKIP
Now we evaluate the right-hand side.
RHS=z1+(z2+z3)
=(a+bi)+((c+di)+(e+fi))
=(a+bi)+(c+e+di+fi)
=(a+bi)+(c+e+(d+f)i)
=a+bi+c+e+(d+f)i
=a+c+e+bi+(d+f)i
=(a+c+e)+(b+d+f)i(2)
SKIP
From (1) and (2)
⇒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) (Associative Property for Multiplication)
(z1⋅z2)⋅z3=z1⋅(z2⋅z3)
LHS=(z1⋅z2)⋅z3
=((a+bi)(c+di))⋅(e+fi)
=((ac−bd)+(ad+bc)i)⋅(e+fi)
=[(ac−bd)e−(ad+bc)f]+[(ac−bd)f+(ad+bc)e]i
=[ace−bde−adf−bcf]+[acf−bdf+ade+bce]i(1)
SKIP
Now we evaluate the right-hand side.
RHS=z1⋅(z2⋅z3)
=(a+bi)⋅((c+di)(e+fi))
=(a+bi)⋅((ce−df)+(cf+de)i)
=[(a)(ce−df)−(b)(cf+de)]+[(a)(cf+de)+(b)(ce−df)]i
=[ace−adf−bcf−bde]+[acf+ade+bce−bdf]i
=[ace−bde−adf−bcf]+[acf−bdf+ade+bce]i(2)
SKIP
From (1) and (2)
⇒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+z1z3 (Distributive Property)