Properties of Multiplication of Complex Numbers

Algebraic Laws for Complex Products

Complex multiplication follows the familiar commutative, associative, identity, and distributive laws. Let $z_1, z_2,$ and $z_3$ be any complex numbers. Each law is verified with one numeric example, so you can follow the expansion of the product yourself.

Commutative Property

The commutative property means that the order in the multiplication of two complex numbers does not affect the result.

z_1 \times z_2 = z_2 \times z_1

Example:

Let $z_1 = 1+2i$ and $z_2 = 3-i$ .

\begin{aligned}z_1 \times z_2 &= (1+2i)(3-i) \\ &= 1(3-i) + 2i(3-i) \\ &= 3-i+6i-2i^2 \\ &= 3+5i-2(-1) \\ &= 5+5i\end{aligned}

Both products equal $5+5i$ , which verifies commutativity in this example.

Associative Property

The associative property states that when multiplying three or more complex numbers, the grouping of the multiplication does not change the result.

(z_1 \times z_2) \times z_3 = z_1 \times (z_2 \times z_3)

Example:

Let $z_1 = i$ , $z_2 = 2$ , and $z_3 = 3-i$ .

\begin{aligned}(z_1 \times z_2) \times z_3 &= (i \times 2) \times (3-i) \\ &= 2i(3-i) \\ &= 6i - 2i^2 \\ &= 6i - 2(-1) \\ &= 2+6i\end{aligned}

Both groupings equal $2+6i$ , which verifies associativity in this example.

Multiplicative Identity

The complex number $1 = 1+0i$ is the identity element for multiplication. This means that any complex number multiplied by $1$ results in the complex number itself.

z \times 1 = z = 1 \times z

Example:

Let $z = 4-7i$ .

\begin{aligned} z \times 1 &= (4-7i)(1+0i) \\ &= 4(1) - (-7)(0) \\ &\quad{}+ i(4(0) + (-7)(1)) \\ &= 4 - 7i \\ &= z \end{aligned}

Distributive Property of Multiplication over Addition

This property connects the operations of multiplication and addition of complex numbers.

\begin{aligned} &z_1 \times (z_2 + z_3) \\ &= (z_1 \times z_2) + (z_1 \times z_3) \end{aligned}

Example:

Let $z_1=2$ , $z_2 = 1+i$ , $z_3 = 3-2i$ .

Left side: $z_1 \times (z_2 + z_3)$

\begin{aligned} &z_1 \times (z_2 + z_3) \\ &= 2 \times ((1+i) + (3-2i)) \end{aligned}

Right side: $(z_1 \times z_2) + (z_1 \times z_3)$

\begin{aligned} (z_1 \times z_2) + (z_1 \times z_3) &= (2(1+i)) \\ &\quad{}+ (2(3-2i)) \end{aligned}

Both sides equal $8-2i$ , which verifies the distributive law in this example.

Example Proof Using Properties

The same laws prove algebraic identities. For any $z_1, z_2$ , expand $(z_1 + z_2)^2$ as follows:

Write the square as a product of two equal factors:

\begin{aligned} &(z_1+z_2)^2 \\ &=(z_1+z_2)(z_1+z_2) \end{aligned}

Apply the distributive property:

=z_1(z_1+z_2) +z_2(z_1+z_2)

Apply the distributive property:

=(z_1z_1+z_1z_2) +(z_2z_1+z_2z_2)

Write repeated equal factors as squares:

\begin{aligned} &=z_1^2+z_1z_2+z_2z_1+z_2^2 \end{aligned}

Use commutativity, $z_2z_1=z_1z_2$ :

=z_1^2+z_1z_2+z_1z_2+z_2^2

Combine like terms:

= z_1^2 + 2z_1z_2 + z_2^2

Multiplicative Inverse

Every non-zero complex number $z = x + iy \neq 0$ has a multiplicative inverse, denoted as $z^{-1}$ or $1/z$ , such that $z \times z^{-1} = 1$ .

Let $z^{-1} = u + iv$ . Then:

z \times z^{-1} = 1

Based on the equality of two complex numbers, we obtain the system of equations:

$xu - yv = 1$
$xv + uy = 0$

To isolate $u$ , multiply the first equation by $x$ and the second by $y$ , then add them:

x^2u-xyv=x

To isolate $v$ , multiply the first equation by $-y$ and the second by $x$ , then add them. Because $z\neq0$ , we have $x^2+y^2>0$ and may divide by this value:

-xyu+y^2v=-y

u = \frac{x}{x^2+y^2}

So, the multiplicative inverse of $z = x+iy$ is:

z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}

Because $x^2+y^2 = |z|^2$ and $x-iy = \bar{z}$ , the inverse formula can also be written as:

z^{-1} = \frac{\bar{z}}{|z|^2}

Algebraic Laws for Complex Products

Commutative Property

The commutative property means that the order in the multiplication of two complex numbers does not affect the result.

z_1 \times z_2 = z_2 \times z_1

Example:

Let $z_1 = 1+2i$ and $z_2 = 3-i$ .

\begin{aligned}z_1 \times z_2 &= (1+2i)(3-i) \\ &= 1(3-i) + 2i(3-i) \\ &= 3-i+6i-2i^2 \\ &= 3+5i-2(-1) \\ &= 5+5i\end{aligned}

Both products equal $5+5i$ , which verifies commutativity in this example.

Associative Property

The associative property states that when multiplying three or more complex numbers, the grouping of the multiplication does not change the result.

(z_1 \times z_2) \times z_3 = z_1 \times (z_2 \times z_3)

Example:

Let $z_1 = i$ , $z_2 = 2$ , and $z_3 = 3-i$ .

\begin{aligned}(z_1 \times z_2) \times z_3 &= (i \times 2) \times (3-i) \\ &= 2i(3-i) \\ &= 6i - 2i^2 \\ &= 6i - 2(-1) \\ &= 2+6i\end{aligned}

Both groupings equal $2+6i$ , which verifies associativity in this example.

Multiplicative Identity

The complex number $1 = 1+0i$ is the identity element for multiplication. This means that any complex number multiplied by $1$ results in the complex number itself.

z \times 1 = z = 1 \times z

Example:

Let $z = 4-7i$ .

\begin{aligned} z \times 1 &= (4-7i)(1+0i) \\ &= 4(1) - (-7)(0) \\ &\quad{}+ i(4(0) + (-7)(1)) \\ &= 4 - 7i \\ &= z \end{aligned}

Distributive Property of Multiplication over Addition

This property connects the operations of multiplication and addition of complex numbers.

\begin{aligned} &z_1 \times (z_2 + z_3) \\ &= (z_1 \times z_2) + (z_1 \times z_3) \end{aligned}

Example:

Let $z_1=2$ , $z_2 = 1+i$ , $z_3 = 3-2i$ .

Left side: $z_1 \times (z_2 + z_3)$

\begin{aligned} &z_1 \times (z_2 + z_3) \\ &= 2 \times ((1+i) + (3-2i)) \end{aligned}

Right side: $(z_1 \times z_2) + (z_1 \times z_3)$

\begin{aligned} (z_1 \times z_2) + (z_1 \times z_3) &= (2(1+i)) \\ &\quad{}+ (2(3-2i)) \end{aligned}

Both sides equal $8-2i$ , which verifies the distributive law in this example.

Example Proof Using Properties

The same laws prove algebraic identities. For any $z_1, z_2$ , expand $(z_1 + z_2)^2$ as follows:

Write the square as a product of two equal factors:

\begin{aligned} &(z_1+z_2)^2 \\ &=(z_1+z_2)(z_1+z_2) \end{aligned}

Apply the distributive property:

=z_1(z_1+z_2) +z_2(z_1+z_2)

Apply the distributive property:

=(z_1z_1+z_1z_2) +(z_2z_1+z_2z_2)

Write repeated equal factors as squares:

\begin{aligned} &=z_1^2+z_1z_2+z_2z_1+z_2^2 \end{aligned}

Use commutativity, $z_2z_1=z_1z_2$ :

=z_1^2+z_1z_2+z_1z_2+z_2^2

Combine like terms:

= z_1^2 + 2z_1z_2 + z_2^2

Multiplicative Inverse

Every non-zero complex number $z = x + iy \neq 0$ has a multiplicative inverse, denoted as $z^{-1}$ or $1/z$ , such that $z \times z^{-1} = 1$ .

Let $z^{-1} = u + iv$ . Then:

z \times z^{-1} = 1

Based on the equality of two complex numbers, we obtain the system of equations:

$xu - yv = 1$
$xv + uy = 0$

To isolate $u$ , multiply the first equation by $x$ and the second by $y$ , then add them:

x^2u-xyv=x

To isolate $v$ , multiply the first equation by $-y$ and the second by $x$ , then add them. Because $z\neq0$ , we have $x^2+y^2>0$ and may divide by this value:

-xyu+y^2v=-y

u = \frac{x}{x^2+y^2}

So, the multiplicative inverse of $z = x+iy$ is:

z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}

Because $x^2+y^2 = |z|^2$ and $x-iy = \bar{z}$ , the inverse formula can also be written as:

z^{-1} = \frac{\bar{z}}{|z|^2}