# 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 âœ• 3 + 1 âœ• 5i + 2i âœ• 3 + 2i âœ• 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 âœ• 3 – 2 âœ• 5) + i(1 âœ• 5 + 2 âœ• 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(cosÎ¸1 + isinÎ¸1) and z2 = r2(cosÎ¸2 + isinÎ¸2) is given by the formula:

z1z2 = r1r2[cos(Î¸1 + Î¸2) + isin(Î¸1 + Î¸2)]

â‡’ z1z2 = r(cosÎ¸ + isinÎ¸), here r = r1r2 and Î¸ = (Î¸1 + Î¸2).

Now, let us multiply the complex numbers z1 = 6(cos190Â° + isin190Â°) and z2 = 5(cos220Â° + isin220Â°) using the above formula..

On multiplying z1 and z2, we get,

z1z2 = 6(cos190Â° + isin190Â°) âœ• 5(cos220Â° + isin220Â°)

= 6 âœ• 5 [(cos190Â°cos220Â° – sin190Â°sin220Â°) + i(sin190Â°cos220Â° + cos190Â°sin220Â°)]

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

= 30(cos410Â° + isin410Â°)

= r(cosÎ¸ + isinÎ¸), where r = 6 âœ• 5 and Î¸ = (190Â° + 220Â°) = 410Â°.

## With Purely Real and Imaginary Numbers

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

c âœ• (a + ib) = ac + ibc

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

(id) âœ• (a + ib) = iad – bd.

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

8 âœ• (5 + 4i) = 8 âœ• 5 + 8 âœ• 4i = 40 + 32i.

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

(-2i) âœ• (5 + 4i) = (-2i) âœ• 5 + (-2i) âœ• 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 â‹… z-1 = z-1 â‹… 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

Multiply (3 + 8i) and (9 + 4i).

Solution:

As we know, the formula for multiplying two complex numbers is:
(a1 + ib1)(a2 + ib2) = (a1a2 – b1b2) + i(a1b2 + a2b1)
Here, (3 + 8i)(9 + 4i)
= (3 âœ• 9 – 8 âœ• 4) + i(3 âœ• 4 + 8 âœ• 9)
= -5 + 84i

Find the square of -2 + 4i.

Solution:

As we know, the formula for multiplying two complex numbers is:
(a1 + ib1)(a2 + ib2) = (a1a2 – b1b2) + i(a1b2 + a2b1)
Here, (-2 + 4i)2 = (-2 + 4i)(-2 + 4i)
= [(-2) âœ• (-2) – 4 âœ• 4] + i[(-2) âœ• 4 + 4 âœ• (-2)]
= (4 – 16) + i(-8 – 8)
= -12 – 16i

Find the multiplicative inverse of 5 + i.

Solution:

As we know, the multiplicative inverse of any complex number z is
z-1 = ${\dfrac{\overline{z}}{\left| z\right| ^{2}}}$
Here, |z| = ${\sqrt{5^{2}+1^{2}}}$ = ${\sqrt{26}}$
â‡’ |z|2 = 26
${\overline{z}}$ = ${\overline{5+i}}$ = 5 – i
Thus, z-1 = ${\dfrac{5-i}{26}}$