Complex Number Multiplication

To multiply two or more complex numbers, we use the distributive property. It is done using the FOIL method, which is also used to multiply two binomials.

On multiplying two complex numbers: z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂, the product obtained is written as:

 z₁z₂ = (a₁ + ib₁)(a₂ + ib₂)

Formula

The general formula to calculate the product of two complex numbers is:

(a₁ + ib₁)(a₂ + ib₂) = (a₁a₂ – b₁b₂) + i(a₁b₂ + a₂b₁)

Let us now multiply two complex numbers: (1 + 2i) and (3 + 5i)

The product is (1 + 2i)(3 + 5i)

Using the FOIL method, we get 1 ✕ 3 + 1 ✕ 5i + 2i ✕ 3 + 2i ✕ 5i

= 3 + 5i + 6i + 10i²

Replacing the value of i² = -1,

3 + 5i + 6i – 10

Adding the like terms, we get (3 – 10) + (5i + 6i)

= -7 + 11i

Instead of multiplying further likewise, by applying the above formula, we can easily obtain the product as:

(1 + 2i)(3 + 5i)

= (1 ✕ 3 – 2 ✕ 5) + i(1 ✕ 5 + 2 ✕ 3)

= -7 + 11i

The numbers 1 + 2i, 3 + 5i, and their product -7 + 11i give the image in the argand plane, as shown.

Similarly, we can multiply the complex numbers represented in the polar form.

In the Polar Form

The multiplication of complex numbers z₁ = r₁(cosθ₁ + isinθ₁) and z₂ = r₂(cosθ₂ + isinθ₂) is given by the formula:

z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + isin(θ₁ + θ₂)]

⇒ z₁z₂ = r(cosθ + isinθ), here r = r₁r₂ and θ = (θ₁ + θ₂).

Now, let us multiply the complex numbers z₁ = 6(cos190° + isin190°) and z₂ = 5(cos220° + isin220°) using the above formula..

On multiplying z₁ and z₂, we get,

z₁z₂ = 6(cos190° + isin190°) ✕ 5(cos220° + isin220°)

= 6 ✕ 5 [(cos190°cos220° – sin190°sin220°) + i(sin190°cos220° + cos190°sin220°)]

= 30[cos(190° + 220°) + isin(190° + 220°)]

= 30(cos410° + isin410°)

= r(cosθ + isinθ), where r = 6 ✕ 5 and θ = (190° + 220°) = 410°.

With Purely Real and Imaginary Numbers

The formula to multiply a complex number z = a + ib with a purely real number c is: 

c ✕ (a + ib) = ac + ibc

Similarly, the formula to multiply a complex number z = a + ib with a purely imaginary number id is:

(id) ✕ (a + ib) = iad – bd.

Now, multiplying 5 + 4i with the real number 8, we get,

8 ✕ (5 + 4i) = 8 ✕ 5 + 8 ✕ 4i = 40 + 32i.

Again, multiplying 5 + 4i with the imaginary number -2i, we get,

(-2i) ✕ (5 + 4i) = (-2i) ✕ 5 + (-2i) ✕ 4i = -10i – 8i²

By substituting the value of i² = -1, the product is -10i + 8.

Properties

Like the real numbers, the complex numbers follow all the properties of multiplication:

1. Closure Property: On multiplying two complex numbers, the product is always a complex number. Thus, for z₁ and z₂, the product is ${z_{1}\cdot z_{2}}$ = (a₁a₂ – b₁b₂) + i(a₁b₂ + a₂b₁), which is a complex number.
2. Commutative Property: For two complex numbers z₁, and z₂, ${z_{1}\cdot z_{2}=z_{2}\cdot z_{1}}$
3. Associative Law: For the given three complex numbers z₁, z₂, and z₃, ${\left( z_{1}\cdot z_{2}\right) \cdot z_{3}}$ = ${z_{1}\cdot \left( z_{2}\cdot z_{3}\right)}$
4. Multiplicative Identity: For any complex number z, there exists 1 (1 + 0i), such that ${z\cdot 1=1\cdot z=z}$
5. Multiplicative Inverse: The multiplicative inverse of any complex number z is denoted by z^-1 and, multiplying with z, gives the product 1, z ⋅ z^-1 = z^-1 ⋅ z = 1.

Here, ${z^{-1}=\dfrac{1}{a+ib}}$ = ${\dfrac{1}{a+ib}\times \dfrac{a-ib}{a-ib}}$ = ${\dfrac{a-ib}{a^{2}+b^{2}}}$ = ${\dfrac{\overline{z}}{\left| z\right| ^{2}}}$

Solved Examples