# Multiplication of Complex Numbers(With Examples)

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Multiplication is one of the four basic operations in math. Multiplication is a complex operation as compared to the addition and subtraction of complex numbers. A complex number is of the form $a + ib$, where $i$ is an imaginary number and $a$ and $b$ are real numbers.

The working mechanism of the multiplication of complex numbers is similar to the multiplication of binomials using the distributive property. In this article, you’ll learn how to multiply complex numbers.

## Multiplication of Two Complex Numbers

Multiplication of two complex numbers is similar to the multiplication of two binomial expressions using the FOIL method.

Mathematically, if we have two complex numbers $z_{1} = a + ib$ and $z_{2} = c + id$, then multiplication of complex numbers $z_{1}$ and $z_{2}$ is written as $z_{1}z_{2} = \left(a + ib \right) \left(c + id \right)$.

Now, we use the FOIL method of multiplication of two binomials. $\left(a + b \right) \left(c + d \right) = ac + ad + bc + bd$.

Therefore, the formula for multiplying complex numbers is given as: $\left(a + ib \right) \left(c + id \right) = ac + iad + ibc + i^{2}bd$

$=> \left(a + ib \right) \left(c + id \right) = \left(ac – bd \right) + i \left(ad + bc \right)$ $\left(\text { Because } i^{2} = -1 \right)$.

Note: The FOIL method:

• The first means that we multiply the terms which occur in the first position of each binomial.
• The outer means that we multiply the terms which are located in both ends (outermost) of the two binomials when written side-by-side.
• The inner means that we multiply the middle two terms of the binomials when written side-by-side.
• The last means that we multiply the terms which occur in the last position of each binomial.
• After obtaining the four (4) partial products coming from the first, outer, inner and last, we simply add them together to get the final answer.

### Examples

Ex 1: Find the product of $3 + 2i$ and $1 + 3i$.

Here, $z_{1} = a + ib = 3 + 2i$ and $z_{2} = c + id = 1 + 3i$.

So, $a = 3$, $b = 2$, $c = 1$ and $d = 3$.

$\left(a + ib \right) \left(c + id \right) = \left(ac – bd \right) + i \left(ad + bc \right)$

$=>\left(3 + 2i \right) \left(1 + 3i \right) = \left(3 \times 1 – 2 \times 3 \right) + i \left(3 \times 3 + 2 \times 1 \right)$

$=>\left(3 + 2i \right) \left(1 + 3i \right) = \left(3 – 6 \right) + i \left(9 + 2 \right)$

$=>\left(3 + 2i \right) \left(1 + 3i \right) = -3 + 11i$

Ex 2: Find the product of $5 – 3i$ and $4 + 2i$.

Here, $z_{1} = a + ib = 5 – 3i$ and $z_{2} = c + id = 4 + 2i$.

So, $a = 5$, $b = -3$, $c = 4$ and $d = 2$.

$\left(a + ib \right) \left(c + id \right) = \left(ac – bd \right) + i \left(ad + bc \right)$

$=>\left(5 – 3i \right) \left(4 + 2i \right) = \left(5 \times 4 – \left(-3 \right) \times 2 \right) + i \left(5 \times 2 + \left(-3 \right) \times 4 \right)$

$=>\left(5 – 3i \right) \left(4 + 2i \right) = \left(20 + 6 \right) + i \left(10 – 12 \right)$

$=>\left(5 – 3i \right) \left(4 + 2i \right) = 26 – 2i$

## Multiplication Of Complex Numbers in Polar Form

A complex number in polar form is written as $z = r \left(\cos \theta + i \sin \theta \right)$, where $r$ is the modulus of the complex number and $\theta$ is its argument. Now, the formula for multiplying complex numbers $z_{1} = r_{1} \left(\cos \theta_{1} + i \sin \theta_{1} \right)$ and $z_{2} = r_{2} \left(\cos \theta_{2} + i \sin \theta_{2} \right)$ in polar form is given as:

$z_{1}z_{2} = (r_{1} \left(\cos \theta_{1} + i \sin \theta_{1} \right))(r_{2} \left( \cos \theta_{2} + i \sin \theta_{2} \right))$

$= r_{1} r_{2} \left(\cos \theta_{1} \cos \theta_{2} + i \cos \theta_{1} \sin \theta_{2} + i \sin \theta_{1} \cos \theta_{2} + i^{2} \sin \theta_{1} \sin \theta_{2} \right)$

$= r_{1} r_{2} \left(\cos \theta_{1} \cos \theta_{2} + i \cos \theta_{1} \sin \theta_{2} + i \sin \theta_{1} \cos \theta_{2} – \sin \theta_{1} \sin \theta_{2} \right)$ $\left( \text { Because } i^{2} = -1 \right)$

$= r_{1} r_{2} \left( \cos \theta_{1} \cos \theta_{2} – \sin \theta_{1} \sin \theta_{2} + i \left( \cos \theta_{1} \sin \theta_{2} + \sin \theta_{1} \cos \theta_{2} \right) \right)$

$= r_1 r_2 \left( \cos \left(\theta_{1} + \theta_{2} \right) + i \sin \left(\theta_{1} + \theta_{2} \right) \right)$ $\left( \text { Because } \cos a \cos b – \sin a \sin b = \cos \left(a + b \right) \text { and } \sin a \cos b + \sin b \cos a = \sin \left(a + b \right) \right)$

Hence the formula for multiplying complex numbers in polar form is $\left( r_{1} \left(\cos \theta_{1} + i \sin \theta_{1} \right) \right) \left(r_{2} \left(\cos \theta_{2} + i \sin \theta_{2} \right) \right) = r_{1} r_{2} \left(\cos \left(\theta_{1} + \theta_{2} \right) + i \sin \left(\theta_{1} + \theta_{2} \right) \right)$.

### Examples

Ex 1: Find the product of $2(\cos \frac {\pi}{6} + i \sin \frac {\pi}{6})$ and $3(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3})$.

Here, $r_{1} = 2$, $r_{2} = 3$, $\theta_{1} = \frac {\pi}{6}$, and $\theta_{2} = \frac {\pi}{3}$

Therefore, $2\left(\cos \frac {\pi}{6} + i \sin \frac {\pi}{6} \right) \times 3\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3} \right)$

$= 2 \times 3 \left(\cos \left(\frac {\pi}{6} + \frac {\pi}{3} \right) + i \sin \left(\frac {\pi}{6} + \frac {\pi}{3} \right) \right)$

$= 6 \left(\cos \frac {\pi}{2} + i \sin \frac {\pi}{2} \right)$

## Properties of Multiplication of Complex Numbers

Following are the properties of the multiplication of complex numbers:

• Closure Property: The product of complex numbers is also a complex number. Hence, it holds the closure property.
• Commutative Property: The multiplication of complex numbers is commutative.
• Associative Property: The multiplication of complex numbers is associative.
• Distributive Property: The multiplication of complex numbers is distributive over addition and subtraction.
• Multipliative Identity: $1 + 0i$ is the multiplicative identity of the complex numbers, i.e., for a complex number $z$, we have $z \times \left(1 + 0i \right) = \left(1 + 0i \right) \times z = z$.
• Multipliative Inverse: For a complex number $z$, the multiplicative inverse in complex numbers is $\frac {1}{z}$, i.e., $z \times \frac {1}{z} = \frac {1}{z} \times z = 1 + 0i$.

## Conclusion

The process of multiplication of two complex numbers is similar to the process of multiplication of two binomial expressions. For multiplication of complex numbers closure property, commutative property, associative property, and distributive property hold, and there exist multiplicative identity and multiplicative inverse for complex numbers.

## Practice Problems

Find the product of the following complex numbers

• $z_{1} = 2 – 3i$, $z_2 = 5 + i$
• $z_{1} = -3 + 2i$, $z_2 = 3 + 4i$
• $z_{1} = 1 – i$, $z_2 = 1 + i$
• $z_{1} = 4 + 2i$, $z_2 = -2 – 3i$
• $z_{1} = 1 – 5i$, $z_2 = 4 + 6i$
• $z_{1} = 3\left(\cos \frac {\pi}{3} + i\sin \frac {\pi}{3}\right)$, $z_{2} = 5\left(\cos \frac {\pi}{2} + i\sin \frac {\pi}{2}\right)$
• $z_{1} = -2\left(\cos \frac {\pi}{4} + i\sin \frac {\pi}{4}\right)$, $z_{2} = 2\left(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6}\right)$

## FAQs

### What is the multiplication of complex numbers formula?

The formula to multiply two complex numbers $z_{1} = a + ib$ and $z_{2} = c + id$ is $\left(a + ib \right) \left(c + id \right) = \left(ac – bd \right) + i \left(ad + bc \right)$.

### What is the multiplication of complex numbers in the polar form formula?

The formula to multiply two complex numbers in polar form $z_{1} = r_{1} \left(\cos \theta_{1} + i \sin \theta_{1} \right)$ and $z_{2} = r_{2} \left(\cos \theta_{2} + i \sin \theta_{2} \right)$ is $r_{1} r_{2} \left(\cos \left(\theta_{1} + \theta_{2} \right) + i \sin \left(\theta_{1} + \theta_{2} \right) \right)$.

### What Happens When You Multiply Two Imaginary Numbers?

When two imaginary numbers are multiplied, the result is a real number, since $i \times i = -1$.