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# Operation with Complex Numbers

So far, we have dealt with real numbers such as:

$2,\frac{1}{3},\sqrt{3},7.09,...$

In this section, we shall look at a new concept called an imaginary number. Consider the square root of 2. We know that this yields the non-repeating decimal

$\sqrt{2}=1.414213562...$

Now, what is the square root of -2? You might think that there is no solution to the square root of a negative number. However, this is not true! In fact, this is where the imaginary number comes into play. The concept of an imaginary number stems from the imaginary unit, denoted by the letter i, and is represented by the following derivation:

${i}^{2}=-1⇔\sqrt{-1}=i$

Thus, the square root of -2 is simply

$\sqrt{-2}=\sqrt{-1×2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-2}=\sqrt{-1}\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-2}=i\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-2}=1.414i\left(correctto3decimalplaces\right)$

As a matter of fact, we can add real and imaginary numbers together. This structure of numbers leads us to the idea of a complex number.

A complex number is an algebraic expression that includes the factor i = √-1 and is written in the form z = a + bi.

## Standard Form of Complex Numbers

The standard form of complex numbers is

$z=a+ib$

where

• Re (z) = a is the real part of the complex number z

• Im (z) = b is the imaginary part of the complex number z

This is also denoted by

$z=Re\left(z\right)+Im\left(z\right)i=a+bi$

### Real and Imaginary Numbers

There are two important subclasses of complex numbers: for a complex number z = a + bi

• If Im (z) = 0, then z = a is a real number

• If Re (z) = 0, then z = bi is said to be purely imaginary

### Why are Complex Numbers Important?

Complex numbers have a range of applications. For instance, they are widely used in the field of electrical engineering and quantum mechanics. Complex numbers also help us solve polynomial equations that do not have any real solutions: have a look at Graph and Solve Quadratic Equations which explains how to do this.

We can conduct basic arithmetic operations with complex numbers such as addition, subtraction, multiplication, and division.

## Operations with Complex Numbers; Addition and Subtraction

In this section we will explain the most important operations you should be able to perform with complex numbers:

• Addition and subtraction of complex numbers
• Scalar multiplication
• Multiplication and division of complex numbers

### Addition and Subtraction of Complex Numbers

To add complex numbers, simply add the corresponding real and imaginary parts. The same rule applies when subtracting complex numbers.

Let z1 and z2 be two complex numbers with z1 = a + bi and z2 = c + di, where a, b, c, and d are real numbers.

#### Addition of Complex Numbers Formula

${z}_{1}+{z}_{2}=\left(a+bi\right)+\left(c+di\right)$

Distributing the positive sign in the second term (to both the real and imaginary parts) and collecting like terms, we obtain

${z}_{1}+{z}_{2}=\left(a+c\right)+\left(b+d\right)i$

#### Subtraction of Complex Numbers Formula

${z}_{1}-{z}_{2}=\left(a+bi\right)-\left(c+di\right)$

Distributing the negative sign in the second term (to both the real and imaginary parts) and collecting like terms, we obtain

${z}_{1}-{z}_{2}=\left(a-c\right)+\left(b-d\right)i$

Let α = 3 - 2i and β = 5 + 7i be two complex numbers

Calculate α + β

$\alpha +\beta =\left(3-2i\right)+\left(5+7i\right)\phantom{\rule{0ex}{0ex}}⇒\alpha +\beta =\left(3+5\right)+\left(-2i+7i\right)\phantom{\rule{0ex}{0ex}}⇒\alpha +\beta =8+5i$

Determine α - β

$\alpha -\beta =\left(3-2i\right)-\left(5+7i\right)\phantom{\rule{0ex}{0ex}}⇒\alpha -\beta =\left(3-5\right)+\left(-2i-7i\right)\phantom{\rule{0ex}{0ex}}⇒\alpha -\beta =-2-9i\phantom{\rule{0ex}{0ex}}⇒\alpha -\beta =-\left(2+9i\right)$

### Scalar Multiplication of Complex Numbers

The Scalar Multiplication of Complex Numbers is the multiplication of a real number and a complex number. In this case, the real number is also called the scalar.

To multiply a complex number by a scalar, simply multiply both the real and imaginary parts by the scalar separately.

Let z = a + bi be a complex number and c be a scalar, where a, b and c are real numbers.

#### Scalar Multiplication of Complex Numbers Formula

$c×z=c\left(a+bi\right)=ca+cbi$

Let α = 3 - 2i and β = 5 + 7i be two complex numbers

Find 7α

In this case, we are multiplying the complex number α by the real number 7 (also called scalar).

$7\alpha =7\left(3-2i\right)\phantom{\rule{0ex}{0ex}}⇒7\alpha =21-14i$

Evaluate 2β

In this case, we are multiplying the complex number β by the real number 2 (also called scalar).

$2\beta =2\left(5+7i\right)\phantom{\rule{0ex}{0ex}}⇒2\beta =10+14i$

## Multiplication of Complex Numbers

Multiplying complex numbers is exactly the same as the binomial expansion technique: apply the FOIL method and combine like terms.

### Multiplication of Complex Numbers Formula

${z}_{1}×{z}_{2}=\left(a+bi\right)\left(c+di\right)=ac-bd+\left(cb+ad\right)i$

This is how the FOIL method works, step-by-step.

Let z1 and z2 be two complex numbers with z1 = a + bi and z2 = c + di, where a, b, c, and d are real numbers. To multiply them

1. Write both in the standard form.
2. Perform the binomial expansion.
3. Combine like terms.
${z}_{1}×{z}_{2}=\left(a+bi\right)\left(c+di\right)\phantom{\rule{0ex}{0ex}}⇒{z}_{1}×{z}_{2}=ac+bci+adi+bd{i}^{2}$

Noting that i2 = -1, we obtain

${z}_{1}×{z}_{2}=ac+bci+adi+bd\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒{z}_{1}×{z}_{2}=ac+bci+adi-bd$

Simplifying this, we get

$ac-bd+\left(cb+ad\right)i$

Let α = 3 - 2i and β = 5 + 7i be two complex numbers.

Find α x β

$\alpha ×\beta =\left(3-2i\right)\left(5+7i\right)\phantom{\rule{0ex}{0ex}}⇒\alpha ×\beta =15+21i-10i-14{i}^{2}\phantom{\rule{0ex}{0ex}}⇒\alpha ×\beta =15+11i-14\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒\alpha ×\beta =15+11i+14\phantom{\rule{0ex}{0ex}}⇒\alpha ×\beta =29+11i$

## Division of Complex Numbers

If you have a fraction of complex numbers, multiply the numerator and denominator by the complex conjugate of the denominator.

For a complex number z = a + bi, the complex conjugate of z is denoted by z* = a - bi.

After that, expand and simplify the expression to the standard form of complex numbers. The result is given by the following formula:

### Division of Complex Numbers Formula

$\frac{{z}_{1}}{{z}_{2}}=\frac{a+bi}{c+di}=\frac{ac+bd+\left(bc-ad\right)i}{{c}^{2}+{d}^{2}}=\frac{ac+bd}{{c}^{2}+{d}^{2}}+\frac{bc-ad}{{c}^{2}+{d}^{2}}i$

When dividing complex numbers be sure to write the final answer in its standard form.

Let's see in practice and step-by-step how to perform complex numbers division. Let z1 and z2 be two complex numbers with z1 = a + bi and z2 = c + di, where a, b, c, and d are real numbers. Dividing z1 by z2, we obtain

$\frac{{z}_{1}}{{z}_{2}}=\frac{a+bi}{c+di}$

The complex conjugate of the denominator, z2 is z2* = c - di.

Now multiplying both the numerator and denominator by z2*, we get

$\frac{{z}_{1}}{{z}_{2}}=\frac{a+bi}{c+di}×\frac{c-di}{c-di}\phantom{\rule{0ex}{0ex}}⇒\frac{{z}_{1}}{{z}_{2}}=\frac{\left(a+bi\right)\left(c-di\right)}{\left(c+di\right)\left(c-di\right)}$

Expanding this expression, we obtain

$\frac{{z}_{1}}{{z}_{2}}=\frac{ac+bci-adi-bd{i}^{2}}{{c}^{2}+cdi-cdi-{d}^{2}{i}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{z}_{1}}{{z}_{2}}=\frac{ac+bci-adi-bd\left(-1\right)}{{c}^{2}+\overline{)cdi}-\overline{)cdi}-{d}^{2}\left(-1\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{{z}_{1}}{{z}_{2}}=\frac{ac+bci-adi+bd}{{c}^{2}+{d}^{2}}$

Finally, combining like terms, we have

$\frac{{z}_{1}}{{z}_{2}}=\frac{ac+bd+\left(bc-ad\right)i}{{c}^{2}+{d}^{2}}$

Let α = 3 - 2i and β = 5 + 7i be two complex numbers. Here, β is the denominator. The complex conjugate of β is β* = 5 - 7i.

Calculate α ÷ β

$\frac{\alpha }{\beta }=\frac{3-2i}{5+7i}$

Here, β is the denominator. The complex conjugate of β is β* = 5 - 7i. Thus, multiplying the numerator and denominator by β* yields:

$\frac{\alpha }{\beta }=\frac{3-2i}{5+7i}×\frac{5-7i}{5-7i}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{\left(3-2i\right)\left(5-7i\right)}{\left(5+7i\right)\left(5-7i\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{15-21i-10i+14{i}^{2}}{25-\overline{)35i}+\overline{)35i}-49{i}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{15-31i+14\left(-1\right)}{25-49\left(-1\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{15-31i-14}{25+49}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{1-31i}{74}\phantom{\rule{0ex}{0ex}}⇒\frac{\alpha }{\beta }=\frac{1}{74}-\frac{31}{74}i$

## Operation with Complex Numbers - Key takeaways

 Operation Formula Addition ${z}_{1}+{z}_{2}=\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$ Subtraction ${z}_{1}-{z}_{2}=\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i$ Scalar Multiplication $c×z=c\left(a+bi\right)=ca+cbi$ Multiplication ${z}_{1}×{z}_{2}=\left(a+bi\right)\left(c+di\right)=ac-bd+\left(cb+ad\right)i$ Division $\frac{{z}_{1}}{{z}_{2}}=\frac{a+bi}{c+di}=\frac{ac+bd}{{c}^{2}+{d}^{2}}+\frac{bc-ad}{{c}^{2}+{d}^{2}}i$

How to do operations with complex numbers: To conduct operations with complex numbers, we must first identify the real and imaginary parts of the complex number.

Operations with complex numbers include addition, subtraction, multiplication and division.

How to solve operations with complex numbers: To solve operations with complex numbers, we must first identify the real part and imaginary part of the complex number and then perform the given arithmetic procedure

How to divide complex numbers:

1. Multiply the numerator and denominator by the complex conjugate
2. Expand and simplify the expression
3. Write the final answer in standard form as a + bi

The rules of a complex number refer to the relationship between a given complex number say z = a + bi and its complex conjugate z* = a - bi.

## Final Operation with Complex Numbers Quiz

Question

Solve (2 + 7i) + (3 − 4i)

5 + 3i

Show question

Question

Solve (6 + 9i) + (12 − 17i)

18 - 8i

Show question

Question

Calculate (9 + 5i) − (4 + 7i)

5 − 2i

Show question

Question

Calculate (2 + 7i) − (13 + 5i)

-11 + 2i

Show question

Question

Given a complex number z = 5 - 2i, find 6z

30 - 12i

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Question

Given a complex number z = 9 + 3i, find -2z

-18 - 6i

Show question

Question

Evaluate (3 + 2i) (5 + 6i)

3 + 28i

Show question

Question

Evaluate (5 - 9i) (4 + 2i)

38 -26 i

Show question

Question

What is the complex conjugate of -2 + i?

-2 - i

Show question

Question

What is the complex conjugate of 15 - 4i?

15 + 4i

Show question

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