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Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Identities

4.3.7

Identities

The Core Identity: (z1+z2)2(z_1 + z_2)^2

The first thing to understand is that the algebraic identities you know from real numbers are not just a coincidence. They come from the fundamental laws of arithmetic — the distributive law, the commutative law of multiplication, and the definition of squaring. Since complex numbers obey all these same laws, every identity that holds for real numbers also holds for complex numbers.

The section begins by proving the square of a sum:

(z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2

Proof.

Start with the definition of squaring: (z1+z2)2(z_1 + z_2)^2 means (z1+z2)(z1+z2)(z_1 + z_2)(z_1 + z_2).

Apply the distributive law (treating the first bracket as a single expression):

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

Now apply the distributive law again to each term:

=z1z1+z2z1+z1z2+z2z2= z_1z_1 + z_2z_1 + z_1z_2 + z_2z_2

Use the commutative law of multiplication (z2z1=z1z2z_2z_1 = z_1z_2):

=z12+z1z2+z1z2+z22= z_1^2 + z_1z_2 + z_1z_2 + z_2^2

Combine the two middle terms:

=z12+2z1z2+z22= z_1^2 + 2z_1z_2 + z_2^2

That's the proof. Notice that the only properties used are the distributive law and the commutative law — both of which hold for complex numbers exactly as they do for reals.

Watch out

The step z2z1=z1z2z_2z_1 = z_1z_2 uses commutativity of multiplication. This is true for complex numbers, but be careful — in other number systems (like matrices) this step would fail, and the identity would not hold.

Four Derived Identities

Once the square identity is established, the textbook lists four more identities that can be proved in the same way. Each one is a direct translation of a real-number identity into the complex setting.

(i) Square of a Difference

(z1−z2)2=z12−2z1z2+z22(z_1 - z_2)^2 = z_1^2 - 2z_1z_2 + z_2^2

Proof.

Write (z1−z2)2=(z1−z2)(z1−z2)(z_1 - z_2)^2 = (z_1 - z_2)(z_1 - z_2). Expand using the distributive law:

=(z1−z2)z1−(z1−z2)z2= (z_1 - z_2)z_1 - (z_1 - z_2)z_2

=z12−z2z1−z1z2+z22= z_1^2 - z_2z_1 - z_1z_2 + z_2^2

=z12−z1z2−z1z2+z22(commutativity)= z_1^2 - z_1z_2 - z_1z_2 + z_2^2 \quad (\text{commutativity})

=z12−2z1z2+z22= z_1^2 - 2z_1z_2 + z_2^2

Tip

A faster way: replace z2z_2 by −z2-z_2 in the sum identity. Since (z1+(−z2))2=z12+2z1(−z2)+(−z2)2=z12−2z1z2+z22(z_1 + (-z_2))^2 = z_1^2 + 2z_1(-z_2) + (-z_2)^2 = z_1^2 - 2z_1z_2 + z_2^2, you get the same result without redoing the expansion.

(ii) Cube of a Sum

(z1+z2)3=z13+3z12z2+3z1z22+z23(z_1 + z_2)^3 = z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3

Proof.

Write (z1+z2)3=(z1+z2)(z1+z2)2(z_1 + z_2)^3 = (z_1 + z_2)(z_1 + z_2)^2. Use the square identity we already proved:

=(z1+z2)(z12+2z1z2+z22)= (z_1 + z_2)(z_1^2 + 2z_1z_2 + z_2^2)

Now expand term by term:

=z1(z12+2z1z2+z22)+z2(z12+2z1z2+z22)= z_1(z_1^2 + 2z_1z_2 + z_2^2) + z_2(z_1^2 + 2z_1z_2 + z_2^2)

=z13+2z12z2+z1z22+z2z12+2z1z22+z23= z_1^3 + 2z_1^2z_2 + z_1z_2^2 + z_2z_1^2 + 2z_1z_2^2 + z_2^3

Group like terms (using commutativity to combine z12z2z_1^2z_2 and z2z12z_2z_1^2, and z1z22z_1z_2^2 and 2z1z222z_1z_2^2):

=z13+(2z12z2+z12z2)+(z1z22+2z1z22)+z23= z_1^3 + (2z_1^2z_2 + z_1^2z_2) + (z_1z_2^2 + 2z_1z_2^2) + z_2^3

=z13+3z12z2+3z1z22+z23= z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3

(iii) Cube of a Difference

(z1−z2)3=z13−3z12z2+3z1z22−z23(z_1 - z_2)^3 = z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3

Proof.

Replace z2z_2 by −z2-z_2 in the cube-of-sum identity:

(z1+(−z2))3=z13+3z12(−z2)+3z1(−z2)2+(−z2)3(z_1 + (-z_2))^3 = z_1^3 + 3z_1^2(-z_2) + 3z_1(-z_2)^2 + (-z_2)^3

=z13−3z12z2+3z1z22−z23= z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3

Alternatively, expand (z1−z2)3=(z1−z2)(z1−z2)2(z_1 - z_2)^3 = (z_1 - z_2)(z_1 - z_2)^2 using the square-of-difference identity — you'll get the same result.

(iv) Difference of Squares

z12−z22=(z1−z2)(z1+z2)z_1^2 - z_2^2 = (z_1 - z_2)(z_1 + z_2)

Proof.

Start from the right-hand side:

(z1−z2)(z1+z2)=z1(z1+z2)−z2(z1+z2)(z_1 - z_2)(z_1 + z_2) = z_1(z_1 + z_2) - z_2(z_1 + z_2)

=z12+z1z2−z2z1−z22= z_1^2 + z_1z_2 - z_2z_1 - z_2^2

=z12+z1z2−z1z2−z22(commutativity)= z_1^2 + z_1z_2 - z_1z_2 - z_2^2 \quad (\text{commutativity})

=z12−z22= z_1^2 - z_2^2

Important

This is the same identity you know from real numbers: a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b). It works for complex numbers because the only operations involved are addition, subtraction, and multiplication — all of which behave identically for complex numbers.

The Bigger Picture

The textbook makes an important observation after listing these four identities: …

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