Multiplying Complex Numbers

The product of two complex numbers, \( z_1 = (a, b) \) and \( z_2 = (c, d) \), is calculated using the following formula: $$ (a,b) \cdot (c,d) = (a \cdot c - b \cdot d, a \cdot d + b \cdot c ) $$

When complex numbers are written in algebraic form as \( z_1 = a + bi \) and \( z_2 = c + di \), their product expands as: $$ (a+bi) \cdot (c+di) = a \cdot (c+di) + bi \cdot (c+di) $$ If the numbers are expressed in trigonometric form: $$ z_1 = m_1 \cdot [\cos(\gamma_1) + i \cdot \sin(\gamma_1)] $$ $$ z_2 = m_2 \cdot [\cos(\gamma_2) + i \cdot \sin(\gamma_2)] $$ their product follows the identity: $$ z_1 \cdot z_2 = (m_1 \cdot m_2) \cdot [ \cos(\gamma_1 + \gamma_2) + i \cdot \sin(\gamma_1 + \gamma_2) ] $$ Similarly, in exponential form: $$ z_1 = m_1 \cdot e^{i\gamma_1}, \quad z_2 = m_2 \cdot e^{i\gamma_2} $$ we apply the multiplication rule: $$ z_1 \cdot z_2 = (m_1 \cdot m_2) \cdot e^{i(\gamma_1+\gamma_2)} $$

Understanding the Multiplication Process

When multiplying two complex numbers in algebraic form, we treat the operation as if we were multiplying two binomials.

$$ (a+bi) \cdot (c+di) $$

Expanding the expression using standard algebraic rules:

$$ (a+bi) \cdot (c+di) = ac + adi + bci + bdi^2 $$

Since \( i^2 = -1 \), we substitute:

$$ = ac + adi + bci - bd $$

Rearranging the terms, we get:

$$ = (ac - bd) + i(ad + bc) $$

This confirms the formula we introduced at the beginning for multiplying complex numbers.

A Practical Example

Let's take two complex numbers:

$$ z_1 = (3,4) $$ $$ z_2 = (2,1) $$

Visually, they are represented as follows:

Using the multiplication formula \( (a,b)(c,d) = (ac - bd, ad + bc) \), we get:

$$ z_1 \cdot z_2 = (3,4) \cdot (2,1) = (3 \cdot 2 - 4 \cdot 1, 3 \cdot 1 + 4 \cdot 2) = (2,11) $$

So, the product of the two complex numbers is \( (2,11) \).

Example 2

Now, let's express the same complex numbers in algebraic form:

$$ z_1 = 3+4i $$ $$ z_2 = 2+i $$

We multiply them using standard algebraic rules:

$$ z_1 \cdot z_2 = (3+4i) \cdot (2+i) $$

Expanding the expression:

$$ = 3 \cdot (2+i) + 4i \cdot (2+i) $$

$$ = 6 + 3i + 8i + 4i^2 $$

$$ = 6 + 11i + 4i^2 $$

Since we know that \( i^2 = -1 \), we substitute:

$$ = 6 + 11i + 4(-1) $$

$$ = 6 + 11i - 4 $$

$$ = 2 + 11i $$

The result is the same: \( 2 + 11i = (2,11) \).

Example 3

Now, let's express the same numbers in trigonometric form:

$$ z_1 = 5 \cdot ( \cos 53.13^\circ + i \cdot \sin 53.13^\circ ) $$

$$ z_2 = 2.24 \cdot ( \cos 26.57^\circ + i \cdot \sin 26.57^\circ ) $$

Using the trigonometric multiplication rule:

$$ z_1 \cdot z_2 = 5 \cdot ( \cos 53.13^\circ + i \cdot \sin 53.13^\circ ) \cdot 2.24 \cdot ( \cos 26.57^\circ + i \cdot \sin 26.57^\circ ) $$

Applying the angle sum formula:

$$ z_1 \cdot z_2 = 5 \cdot 2.24 \cdot [ \cos (53.13^\circ + 26.57^\circ) + i \cdot \sin (53.13^\circ + 26.57^\circ) ] $$

$$ z_1 \cdot z_2 = 11.02 \cdot [ \cos 79.7^\circ + i \cdot \sin 79.7^\circ ] $$

Example 4

Finally, let's rewrite the numbers in exponential form:

$$ z_1 = 5 \cdot e^{i \cdot 53.13^\circ} $$

$$ z_2 = 2.24 \cdot e^{i \cdot 26.57^\circ} $$

Using the properties of exponentiation:

$$ z_1 \cdot z_2 = 5 \cdot e^{i \cdot 53.13^\circ} \cdot 2.24 \cdot e^{i \cdot 26.57^\circ} $$

Applying the exponent sum rule:

$$ z_1 \cdot z_2 = 11.02 \cdot e^{i \cdot (53.13^\circ + 26.57^\circ)} $$

$$ z_1 \cdot z_2 = 11.02 \cdot e^{i \cdot 79.7^\circ} $$

Properties of Complex Number Multiplication

The multiplication of complex numbers satisfies the following properties:

• Commutative Property
  The order of multiplication does not affect the result: $$ z_1 \cdot z_2 = z_2 \cdot z_1 $$

  Example: Let's multiply two complex numbers: \( z_1 = 1 + 2i \) and \( z_2 = 3 + i \) $$ z_1 \cdot z_2 = (1 + 2i) \cdot (3 + i) $$ $$ z_1 \cdot z_2 = 1 + 7i $$ Now, reversing the order: $$ z_2 \cdot z_1 = (3 + i) \cdot (1 + 2i) $$ $$ z_2 \cdot z_1 = 1 + 7i $$ Since the result remains the same, the commutative property is verified.

• Associative Property
  The associative property states that for any three complex numbers \( z_1 \), \( z_2 \), and \( z_3 \), the way they are grouped does not change the result: $$ (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3) $$ This guarantees that multiplication can be performed in any grouping without affecting the outcome.

  Example: Consider three complex numbers: \( z_1 = 1 + 2i \), \( z_2 = 3 + i \), and \( z_3 = 2 - i \). Computing the product: $$ (z_1 \cdot z_2) \cdot z_3 = [(1 + 2i) \cdot (3 + i)] \cdot (2 - i) $$ $$ (z_1 \cdot z_2) \cdot z_3 = (1 + 7i) \cdot (2 - i) $$ $$ (z_1 \cdot z_2) \cdot z_3 = 9 + 13i $$ Now, changing the grouping: $$ z_1 \cdot (z_2 \cdot z_3) = (1 + 2i) \cdot [(3 + i) \cdot (2 - i)] $$ $$ z_1 \cdot (z_2 \cdot z_3) = (1 + 2i) \cdot (7 - i) $$ $$ z_1 \cdot (z_2 \cdot z_3) = 9 + 13i $$ Since the result is unchanged, the associative property holds.

• Distributive Property
  The product of a complex number \( z_1 \) and a sum \( z_2 + z_3 \) equals the sum of the individual products: $$ z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3 $$

  Example: Consider three complex numbers: \( z_1 = 1 + 2i \), \( z_2 = 3 + i \), and \( z_3 = 2 - i \). First, compute \( z_1 (z_2 + z_3) \): $$ z_1 \cdot (z_2 + z_3) = (1 + 2i) \cdot [(3 + i) + (2 - i)] $$ $$ z_1 \cdot (z_2 + z_3) = (1 + 2i) \cdot 5 $$ $$ z_1 \cdot (z_2 + z_3) = 5 + 10i $$ Now, computing \( z_1 \cdot z_2 + z_1 \cdot z_3 \): $$ z_1 \cdot z_2 + z_1 \cdot z_3 = (1 + 2i) \cdot (3 + i) + (1 + 2i) \cdot (2 - i) $$ $$ = (1 + 7i) + (4 + 3i) $$ $$ = 5 + 10i $$ Since both results match, the distributive property is confirmed.

• Multiplicative Identity
  The multiplicative identity is \( (1,0) \) because multiplying any complex number by \( (1,0) \) leaves it unchanged.

  Example: Multiplying \( z_1 = 1 + 2i \) by \( z_2 = 1 + 0i \): $$ z_1 \cdot z_2 = (1 + 2i) \cdot (1 + 0i) $$ $$ z_1 \cdot z_2 = 1 + 2i $$ The result remains the same, confirming the identity property.

• Multiplication by Zero
  The multiplicative zero property states that multiplying any complex number by \( (0,0) \) results in \( (0,0) \).

  Example: Multiplying \( z_1 = 1 + 2i \) by \( z_2 = 0 + 0i \): $$ z_1 \cdot z_2 = (1 + 2i) \cdot (0 + 0i) $$ $$ z_1 \cdot z_2 = 0 $$ This confirms the property.

• Multiplication of Purely Real Complex Numbers
  When two complex numbers have no imaginary part, meaning they are of the form \( (a,0) \) and \( (b,0) \), their product is simply the product of their real parts: $$ (a, 0) \cdot (b, 0) = (ab, 0) $$

  Example: Consider \( z_1 = (3,0) \) and \( z_2 = (4,0) \). Computing their product: $$ (3,0) \cdot (4,0) = (3 \cdot 4, 0) = (12,0) $$ This confirms that the product of two purely real complex numbers is simply the product of their real parts.

Proofs

A] Commutative Property of Multiplication

To prove that complex number multiplication is commutative, we must show that:

$$ z_1 \cdot z_2 = z_2 \cdot z_1 $$

Let \( z_1 = a + bi \) and \( z_2 = c + di \), where \( a, b, c, d \) are real numbers. We compute the product in both orders:

• First, computing \( z_1 \cdot z_2 \): $$ z_1 \cdot z_2 = (a + bi)(c + di) $$ Expanding using the distributive property: $$ = ac + adi + bci + bdi^2 $$ Since \( i^2 = -1 \), we substitute: $$ = ac - bd + (ad + bc)i $$
• Now, computing \( z_2 \cdot z_1 \): $$ z_2 \cdot z_1 = (c + di)(a + bi) $$ Expanding using the distributive property: $$ = ca + cbi + dai + dbi^2 $$ Again, substituting \( i^2 = -1 \): $$ = ca - db + (cb + ad)i $$

Since \( ac = ca \), \( bd = db \), and \( ad + bc = cb + ad \), the two expressions are identical:

$$ ac - bd + (ad + bc)i = ca - db + (cb + ad)i $$

Thus, we have proven that complex number multiplication is commutative.

B] Associative Property of Multiplication

To prove that complex number multiplication is associative, we must show that for any three complex numbers \( z_1 \), \( z_2 \), and \( z_3 \):

$$ (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3) $$

Let \( z_1 = a + bi \), \( z_2 = c + di \), and \( z_3 = e + fi \), where \( a, b, c, d, e, f \) are real numbers. We compute the product in both groupings:

• First, computing \( (z_1 \cdot z_2) \cdot z_3 \): Compute \( z_1 \cdot z_2 \): $$ z_1 \cdot z_2 = (a + bi)(c + di) = ac - bd + (ad + bc)i $$ Now, multiply by \( z_3 \): $$ (z_1 \cdot z_2) \cdot z_3 = [ac - bd + (ad + bc)i] \cdot (e + fi) $$ Expanding using the distributive property: $$ = (ac - bd)e + (ac - bd)fi + (ad + bc)ei + (ad + bc)fi^2 $$ Substituting \( i^2 = -1 \): $$ = ace - bde - (ad + bc)f + [(ac - bd)f + (ad + bc)e]i $$ $$ = ace - bde - adf - bcf + [(ac - bd)f + (ad + bc)e]i $$
• Now, computing \( z_1 \cdot (z_2 \cdot z_3) \): Compute \( z_2 \cdot z_3 \): $$ z_2 \cdot z_3 = (c + di)(e + fi) = ce - df + (cf + de)i $$ Now, multiply by \( z_1 \): $$ z_1 \cdot (z_2 \cdot z_3) = (a + bi) \cdot [ce - df + (cf + de)i] $$ Expanding using the distributive property: $$ = ace - adf - bcf - bde + [(ac - bd)f + (ad + bc)e]i $$

Since both results are identical, we conclude that:

$$ (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3) $$

Thus, the associative property of complex multiplication is proven.

C] Distributive Property Over Addition

To prove that complex number multiplication distributes over addition, we must show that for any complex numbers \( z_1 \), \( z_2 \), and \( z_3 \):

$$ z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3 $$

Let \( z_1 = a + bi \), \( z_2 = c + di \), and \( z_3 = e + fi \), where \( a, b, c, d, e, f \) are real numbers. We compute both sides of the equation:

• First, computing \( z_1 \cdot (z_2 + z_3) \): Compute the sum \( z_2 + z_3 \): $$ z_2 + z_3 = (c + di) + (e + fi) = (c + e) + (d + f)i $$ Now, multiply by \( z_1 \): $$ z_1 \cdot (z_2 + z_3) = (a + bi) \cdot [(c + e) + (d + f)i] $$ Expanding using the distributive property: $$ = a(c + e) + a(d + f)i + bi(c + e) + bi(d + f)i $$ Substituting \( i^2 = -1 \): $$ = [a(c + e) - b(d + f)] + [a(d + f) + b(c + e)]i $$ $$ = [ac + ae - bd - bf] + [ad + af + bc + be]i $$
• Now, computing \( z_1 \cdot z_2 + z_1 \cdot z_3 \): $$ z_1 \cdot z_2 = (a + bi)(c + di) = ac - bd + (ad + bc)i $$ $$ z_1 \cdot z_3 = (a + bi)(e + fi) = ae - bf + (af + be)i $$ Adding the two results: $$ z_1 \cdot z_2 + z_1 \cdot z_3 = [ac + ae - bd - bf] + [ad + af + bc + be]i $$

Since both results match, we conclude that:

$$ z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3 $$

Thus, the distributive property of complex multiplication is proven.

Additional Notes

Here are some additional observations on complex number multiplication.

• Product of a Complex Number and Its Conjugate
  The product of a complex number \( z = a + bi \) and its conjugate \( \overline{z} = a - bi \) is always a real number. Specifically, it equals the sum of the square of the real part \( a \) and the square of the imaginary coefficient \( b \): $$ z \cdot \overline{z} = (a + bi)(a - bi) = a^2 + b^2 $$ This result corresponds to the square of the modulus of \( z \): $$ |z|^2 = a^2 + b^2 $$

  Example: Consider the complex number \( z = 3 + 4i \). Its conjugate is \( \overline{z} = 3 - 4i \). Let's compute their product: $$ z \cdot \overline{z} = (3 + 4i)(3 - 4i) $$ Using the identity for the product of a binomial and its conjugate: $$ z \cdot \overline{z} = 3^2 - (4i)^2 $$ Since \( i^2 = -1 \), we substitute: $$ z \cdot \overline{z} = 9 - 16(-1) = 9 + 16 = 25 $$ Thus, the product of \( z \) and \( \overline{z} \) is \( 25 \), a real number.

And so on.