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# Division of Complex Numbers(With Examples)

August 21, 2022 This post is also available in: हिन्दी (Hindi)

Division as a reverse process of multiplication is a bit more complicated as compared to the multiplication of complex numbers because of an imaginary unit involved in it. In the case of division, you need to find a term by which you can multiply the numerator and the denominator that will eliminate the imaginary part of the denominator. It is similar to the process of rationalizing the denominator to eliminate the irrational number from the denominator.

Let’s learn about the process of division of complex numbers.

## Division of Complex Numbers

The division of complex numbers is similar to the process of division of real numbers. For two complex numbers $z_{1} = a + ib$ and $z_{2} = c + id$, $z_{1} \div z_{2} = \frac {z_{1}}{z_{2}} = \frac {a + ib}{c + id}$.

$\frac {a + ib}{c + id}$ is same as multiplying $a + ib$ by the multiplicative inverse of $c + id$, i.e., $\frac {a + ib}{c + id} = \left(a + ib \right) \times \frac {1}{c + id}$.

For example, $\frac {2 + 3i}{3 + 4i}$ is same as $\left(2 + 3i \right) \times \frac {1}{3 + 4i}$.

## What is Multiplicative Inverse of a Complex Number?

For a complex number $z = a + ib$, $\frac {1}{z} = \frac {1}{z} = \frac {1}{a + ib}$ is called its multiplicative inverse.  You can observe that in the case of the multiplicative inverse of a complex number the imaginary number is in the denominator. To eliminate the imaginary number from the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator.

Note: Conjugate of a complex number is a complex number with opposite sign of imaginary part.

• Conjugate of $a + ib$ is $a – ib$
• Conjugate of $a – ib$ is $a + ib$

So, $\frac {1}{a + ib}$ becomes $\frac {1}{a + ib} \times \frac {a – ib}{a – ib} = \frac {1 \times \left(a – ib \right)}{\left(a + ib \right)\left(a – ib \right)} = \frac {a – ib}{a^{2} – \left(ib\right)^{2}} = \frac {a – ib}{a^{2} + b^{2}}$.

Note:

• Multiplicative inverse of a complex number $a + ib$ is $\frac {a + ib}{a^{2} + b^{2}}$
• Multiplicative inverse of a complex number $a – ib$ is $\frac {a – ib}{a^{2} + b^{2}}$

### Steps For Computing Multiplicative Inverse of a Complex Number

These are the steps in finding the multiplicative inverse of a complex number

Step 1: For a complex number $a + ib$, write the multiplicative inverse in the form $\frac {1}{a + ib}$

Step 2: Find the conjugate of $a + ib$ which is $a – ib$

Step 3: The multiplicate inverse of $a + ib$ is $\frac {\text { Conjugate of } a + ib}{a^{2} + b^{2}}$, i.e., $\frac {a – ib}{a^{2} + b^{2}}$

### Examples

Let’s consider some examples to understand the process of finding the multiplicative inverse of a complex number.

Ex 1: Find the multiplicative inverse of $2 + 3i$

Conjugate of $2 + 3i$ is $2 – 3i$.

Therefore, multiplicative inverse of $2 + 3i$ is $\frac {2 – 3i}{2^{2} + 3^{2}}= \frac {2 – 3i}{13} = \frac {2}{13} – \frac {3}{13}i$.

Ex 2: Find the multiplicative inverse of $1 – 2i$

Conjugate of $1 – 2i$ is $1 + 2i$.

Therefore, multiplicative inverse of $1 – 2i$ is $\frac {1 + 2i}{1^{2} + 2^{2}} = \frac {1 + 2i}{5} = \frac {1}{5} + \frac {2}{5}i$. Maths can be really interesting for kids

## Steps For Division of Complex Numbers

$z_{1} \div z_{2} = \frac {z_{1}}{z_{2}}$, ie., multiplying $z_{1}$ by multiplicative inverse of $z_{2}$.

The steps to divide complex numbers are.

Step 1: Find the multiplicative inverse of the second complex number, i.e., the divisor

Step 2: Multiply the first complex number, i.e., the dividend with the multiplicative inverse of the second complex number, i.e., the divisor

### Examples

Ex 1: Divide $3 – 7i$ by $2 + 5i$

$\left (3 – 7i \right) \div \left(2 + 5i \right)$

Multiplicative inverse of $2 + 5i$ is $\frac {2 – 5i}{2^{2} + 5^{2}} = \frac {2 – 5i}{29} =\frac {2}{29} – \frac {5}{29}i$.

Therefore, $\left (3 – 7i \right) \div \left(2 + 5i \right) = \left (3 – 7i \right) \times \frac {2 – 5i}{29}$

Now, use the multiplication formula of complex numbers to multiply $3 – 7i$ and $2 – 5i$.

$\left(3 – 7i \right)\times \left(2 – 5i \right) = \left(3 \times 2 – \left(-7 \right) \times \left(-5 \right) \right) + i \left(3 \times \left(-5 \right) + \left(-7 \right) \times 2 \right)$

$= \left(6 – 35 \right) + i \left(-15 + \left(-14 \right) \right) = -29 – 29i$

Therefore, $\left (3 – 7i \right) \div \left(2 + 5i \right) = \frac {-29 – 29i}{29} = -1 – i$

Note: Multiplication formula for two complex numbers $a + ib$ and $c + id$ is $\left(a + ib \right) \left(c + id \right) = \left(ac – bd \right) + i \left(ad + bc \right)$.

Ex 2: Divide $6 + 3i$ by $4 – 8i$

$\left (6 + 3i \right) \div \left(4 – 8i \right)$

Multiplicative inverse of $4 – 8i$ is $\frac {4 + 8i}{4^{2} + 8^{2}} = \frac {4 + 8i}{80} = \frac {1 + 2i}{20} =\frac {1}{20} – \frac {1}{10}i$.

Therefore, $\left (6 + 3i \right) \div \left(4 – 8i \right) = \left (6 + 3i \right) \times \frac {1 + 2i}{20}$

$\left(6 + 3i \right) \times \left(1 + 2i \right) = \left(6 \times 1 – 3 \times 2 \right) + i \left(6 \times 2 + 3 \times 1 \right) = \left(6 – 6 \right) + i \left(12 + 3 \right) = 15i$.

Therefore, $\left (6 + 3i \right) \div \left(4 – 8i \right) = \frac {15i}{20} = \frac {3}{4}i$.

## Division of Complex Numbers in Polar Form

For any two complex numbers $z_{1} = r_{1}\left(\cos \theta_{1} + i\sin \theta_{1} \right)$ and $z_{2} = r_{2}\left(\cos \theta_{2} + i\sin \theta_{2} \right)$, $\frac {z_{1}}{z_{2}}$ is calculated as

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

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

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

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

$= r\left(\cos \theta + i \sin \theta \right)$

where, $r = \frac {r_{1}}{r_{2}}$ and $\theta = \theta_{1} – \theta_{2}$

### Examples

Let’s consider some examples to understand the process of division of two complex numbers in polar form.

Ex 1: Divide $4\left(\cos \frac {\pi}{2} + i \sin \frac {\pi}{2} \right)$ by $2\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3} \right)$

Here, $r_{1} = 4$ and $r_{2} = 2$ and $\theta_{1} = \frac {\pi}{2}, \theta_{2} = \frac {\pi}{3}$

Therefore, $4\left(\cos \frac {\pi}{2} + i \sin \frac {\pi}{2} \right) \div 2\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3} \right)$

$= \frac {4}{2} \left(\cos\left(\frac {\pi}{2} – \frac {\pi}{3} \right) + i\sin\left(\frac {\pi}{2} – \frac {\pi}{3} \right)\right)$

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

## Properties of Division of Complex Numbers

Following are the properties of the division of complex numbers:

• Closure Property: The quotient of complex numbers is also a complex number. Hence, it holds the closure property.
• Commutative Property: The division of complex numbers is not commutative.
• Associative Property: The division of complex numbers is not associative.
• Distributive Property: The division of complex numbers is not distributive over addition and subtraction.

## Conclusion

The division of two complex numbers involves finding the multiplicative inverse of the second complex number, i.e., the divisor, and then multiplying the first complex number i.e., the dividend with the multiplicative inverse of the second complex number i.e., the divisor.

## Practice Problems

1. Find the multiplicative inverse of the following complex numbers
• $2 – 6i$
• $-5 + 4i$
• $12 + 3i$
• $-3 – 5i$
• $-4 + 7i$
2. Find the product of the following pairs of complex numbers
• $7 + 3i$ and $2 – 5i$
• $1 + i$ and $1 – i$
• $15 – 8i$ and $3 – 7i$
• $-9 + 7i$ and $4 + 2i$
• $11 + 6i$ and $3 – 9i$
3. Find the product of the following pairs of complex numbers
• $2\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3}\right) \div 4\left(\cos \frac {\pi}{4} + i \sin \frac {\pi}{4}\right)$
• $5\left(\cos \frac {\pi}{2} + i \sin \frac {\pi}{2}\right) \div 2\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3}\right)$
• $6\left(\cos \frac {\pi}{3} + i \sin \frac {\pi}{3}\right) \div 4\left(\cos \frac {\pi}{6} + i \sin \frac {\pi}{6}\right)$

## FAQs

### Is division applicable for complex numbers?

Yes, the division is applicable for complex numbers. For two complex numbers $z_{1}$ and $z_{2}$, you can divide $z_{1}$ by $z_{2}$ i.e., $\frac {z_{1}}{z_{2}}$ or divide $z_{2}$ by $z_{1}$ i.e., $\frac {z_{2}}{z_{1}}$.

### What is complex division?

Complex division means dividing complex numbers. For two complex numbers $z_{1}$ and $z_{2}$, you can divide $z_{1}$ by $z_{2}$ i.e., $\frac {z_{1}}{z_{2}}$ or divide $z_{2}$ by $z_{1}$ i.e., $\frac {z_{2}}{z_{1}}$.

### What is the multiplicative inverse of a complex number?

Multiplicative inverse of a complex number $z$ is $\frac {1}{z}$. If $z = a + ib$, then its multiplicative inverse is $\frac {a – ib}{a^{2} + b^{2}}$.

### How do you divide and simplify complex numbers?

To divide $z_{1}$ by $z_{2}$, you multiply $z_{1}$ by the multiplicative inverse of $z_{2}$, i.e., by $\frac {1}{z_{2}}$. Therefore, $\frac {z_{1}}{z_{2}}$ is same as $z_{1} \times \frac {1}{z_{2}}$.

Multiplicative inverse of $a + ib$ is given by $\frac {a – ib}{a^{2} + b^{2}}$.