Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Multiplication of Two Complex Numbers

4.3.3

Multiplication of Two Complex Numbers

The Definition of Multiplication

When you multiply two complex numbers, you treat them like binomials in $i$ , but with one crucial rule: $i^2 = -1$ . This single substitution is what makes complex multiplication different from ordinary binomial multiplication.

Let $z_1 = a + ib$ and $z_2 = c + id$ be any two complex numbers. Their product $z_1 z_2$ is defined as:

z_1 z_2 = (a + ib)(c + id)

Expand the product just as you would $(a + x)(c + x)$ :

= ac + iad + ibc + i^2 bd

Now apply $i^2 = -1$ :

= ac + iad + ibc - bd

Group the real terms and the imaginary terms:

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

This is the fundamental formula for multiplying two complex numbers.

(a + ib)(c + id) = (ac - bd) + i(ad + bc)

Example: Multiply $(3 + i5)$ and $(2 + i6)$ .

Here $a = 3$ , $b = 5$ , $c = 2$ , $d = 6$ .

Real part: $ac - bd = 3 \times 2 - 5 \times 6 = 6 - 30 = -24$

Imaginary part: $ad + bc = 3 \times 6 + 5 \times 2 = 18 + 10 = 28$

So $(3 + i5)(2 + i6) = -24 + i28$ .

Watch out

A common mistake is to forget the minus sign in the real part. The term $-bd$ comes from $i^2 bd = -bd$ , not $+bd$ . Always write $i^2 = -1$ explicitly in your expansion until it becomes automatic.

Properties of Multiplication

Complex multiplication behaves in ways that are very familiar from real numbers. The textbook lists six properties, each stated without proof. Here we provide the complete reasoning for each.

(i) The Closure Law

Statement: The product of two complex numbers is always a complex number. For all complex numbers $z_1$ and $z_2$ , $z_1 z_2$ is a complex number.

Why it holds: If $z_1 = a + ib$ and $z_2 = c + id$ , then $z_1 z_2 = (ac - bd) + i(ad + bc)$ . Since $a, b, c, d$ are real numbers, $ac - bd$ and $ad + bc$ are also real numbers. Therefore the result is of the form $(\text{real}) + i(\text{real})$ , which is exactly the definition of a complex number. The product never "escapes" the set of complex numbers.

Note

This property is not trivial — it tells us that complex numbers are closed under multiplication, just as integers are closed under addition. You will never get something like "infinity" or an undefined result from multiplying two complex numbers.

(ii) The Commutative Law

Statement: For any two complex numbers $z_1$ and $z_2$ , $z_1 z_2 = z_2 z_1$ .

Proof: Let $z_1 = a + ib$ and $z_2 = c + id$ .

Compute $z_1 z_2$ :

z_1 z_2 = (ac - bd) + i(ad + bc)

Now compute $z_2 z_1$ :

z_2 z_1 = (c + id)(a + ib) = ca + icb + ida + i^2 db = (ca - db) + i(cb + da)

Since multiplication of real numbers is commutative ( $ca = ac$ , $db = bd$ , $cb = bc$ , $da = ad$ ), we have:

z_2 z_1 = (ac - bd) + i(ad + bc)

This is exactly the same as $z_1 z_2$ . Therefore $z_1 z_2 = z_2 z_1$ .

Tip

Because multiplication is commutative, you can multiply complex numbers in any order. This is useful when simplifying expressions — you can rearrange factors freely.

(iii) The Associative Law

Statement: For any three complex numbers $z_1, z_2, z_3$ , $(z_1 z_2) z_3 = z_1 (z_2 z_3)$ .

Proof: Let $z_1 = a + ib$ , $z_2 = c + id$ , $z_3 = e + if$ .

First compute $(z_1 z_2) z_3$ :

Step 1: $z_1 z_2 = (ac - bd) + i(ad + bc)$ . Call this $p + iq$ where $p = ac - bd$ and $q = ad + bc$ .

Step 2: Multiply $(p + iq)$ by $z_3 = e + if$ :

(p + iq)(e + if) = (pe - qf) + i(pf + qe)

Substitute $p$ and $q$ :

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

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

Now compute $z_1 (z_2 z_3)$ :

Step 1: $z_2 z_3 = (c + id)(e + if) = (ce - df) + i(cf + de)$ . Call this $r + is$ where $r = ce - df$ and $s = cf + de$ .

Step 2: Multiply $z_1 = a + ib$ by $(r + is)$ :

(a + ib)(r + is) = (ar - bs) + i(as + br)

Substitute $r$ and $s$ :

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

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

Compare the two results. The real parts are identical: both are $ace - bde - adf - bcf$ . The imaginary parts are identical: both are $acf - bdf + ade + bce$ . Therefore $(z_1 z_2) z_3 = z_1 (z_2 z_3)$ .

Note

The algebra looks messy, but the key insight is that associativity follows from the associativity of real number multiplication and addition. The $i^2 = -1$ rule does not break this property.

(iv) The Multiplicative Identity

Statement: There exists a complex number $1 + i0$ (denoted simply as $1$ ) such that for every complex number $z$ , $z \cdot 1 = z$ . This number is called the multiplicative identity.

Proof: Let $z = a + ib$ . Then:

z \cdot 1 = (a + ib)(1 + i0)

Using the multiplication formula with $c = 1$ , $d = 0$ :

= (a \cdot 1 - b \cdot 0) + i(a \cdot 0 + b \cdot 1)

= (a - 0) + i(0 + b)

= a + ib = z

So multiplying any complex number by $1$ leaves it unchanged. The number $1 + i0$ behaves exactly like the real number $1$ in multiplication.

Important

The multiplicative identity is the complex number whose real part is $1$ and imaginary part is $0$ . It is the same as the real number $1$ , but written in complex form.

(v) The Multiplicative Inverse

Statement: For every non-zero complex number $z = a + ib$ (where at least one of $a$ or $b$ is non-zero), there exists a complex number called the multiplicative inverse of $z$ , denoted by $\frac{1}{z}$ or $z^{-1}$ , such that $z \cdot \frac{1}{z} = 1$ (the multiplicative identity).

The multiplicative inverse is given by:

\frac{1}{z} = \frac{a}{a^2 + b^2} + i\left(\frac{-b}{a^2 + b^2}\right)

Derivation: We want to find a complex number $x + iy$ such that:

(a + ib)(x + iy) = 1 + i0

Expand the left side:

(ax - by) + i(ay + bx) = 1 + i0

For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. This gives us a system of two equations:

ax - by = 1 \quad \text{(1)}

ay + bx = 0 \quad \text{(2)}

From equation (2): $ay = -bx$ , so $y = -\frac{b}{a}x$ (provided $a \neq 0$ ).

Substitute into equation (1):

ax - b\left(-\frac{b}{a}x\right) = 1

ax + \frac{b^2}{a}x = 1

x\left(a + \frac{b^2}{a}\right) = 1

x\left(\frac{a^2 + b^2}{a}\right) = 1

x = \frac{a}{a^2 + b^2}

Then $y = -\frac{b}{a} \cdot \frac{a}{a^2 + b^2} = \frac{-b}{a^2 + b^2}$ .

Therefore:

\frac{1}{z} = \frac{a}{a^2 + b^2} + i\left(\frac{-b}{a^2 + b^2}\right)

Watch out

The multiplicative inverse exists only when $z \neq 0$ . If $a = 0$ and $b = 0$ , then $a^2 + b^2 = 0$ and the inverse would involve division by zero, which is undefined. Zero has no multiplicative inverse.

Verification: Multiply $z$ by its inverse:

(a + ib)\left(\frac{a}{a^2 + b^2} - i\frac{b}{a^2 + b^2}\right)

Real part: $a \cdot \frac{a}{a^2 + b^2} - b \cdot \left(-\frac{b}{a^2 + b^2}\right) = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$

Imaginary part: $a \cdot \left(-\frac{b}{a^2 + b^2}\right) + b \cdot \frac{a}{a^2 + b^2} = -\frac{ab}{a^2 + b^2} + \frac{ab}{a^2 + b^2} = 0$

So the product is $1 + i0 = 1$ , as required.

Tip

A quick way to find the multiplicative inverse: write $z = a + ib$ , then $\frac{1}{z} = \frac{a - ib}{a^2 + b^2}$ . This is because $(a + ib)(a - ib) = a^2 + b^2$ (a real number), so dividing both sides by $a^2 + b^2$ gives the inverse. The number $a - ib$ is called the complex conjugate of $z$ , which you will study in detail later.

(vi) The Distributive Law …

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