# 2. Basic Operations with Complex Numbers

by M. Bourne

Addition and subtraction of complex numbers works in a similar way to that of adding and subtracting surds. This is not surprising, since the imaginary number
*j* is defined as `j=sqrt(-1)`.

**Addition of Complex Numbers**

Add real parts, add imaginary parts.

**Subtraction of Complex Numbers**

Subtract real parts, subtract imaginary parts.

### Example 1- Addition & Subtraction

**a. **`(6 + 7j) + (3 − 5j) =`

`(6 + 3) + (7 − 5)j = 9 + 2j`

**b. **`(12 + 6j) − (4 + 5j) =`

`(12 − 4) + (6 − 5)j = 8 + j`

## Multiplication of Complex Numbers

Expand brackets as usual, but care with `j^2`!

### Example 2 - Multiplication

Multiply the following.

**a. **`5(2 + 7j)`

Answer

`5(2 + 7j) = 10 + 35j`

**b. **`(6 − j)(5j)`

Answer

`(6 − j)(5j)`

`= 30j − 5j^2`

`= 30j − 5(−1) ``= 5 + 30j`

Easy to understand math videos:

MathTutorDVD.com

**c. **`(2 − j)(3 + j)`

Answer

`(2 − j)(3 + j)`

`= 6 − 3j + 2j − j^2`

`= 6 − j − (−1)`

`= 6 − j + 1 `

`= 7 − j`

Please support IntMath!

**d**. `(5 + 3j)^2`^{ }

Answer

We apply the algebraic expansion `(a+b)^2 = a^2 + 2ab + b^2` as follows:

`(5 + 3j)^2 = 25 + 2(5)(3)j + 9(j^2)`

`= 25 + 30j + 9(-1)`

`= 25 + 30j - 9`

`= 16 + 30j`

**e.** `(2sqrt(-9)-3)(3sqrt(-16)-1)`

Answer

`(2sqrt(-9)-3)(3sqrt(-16)-1)`

`=(2j(3)-3)(3j(4)-1)`

`=(6j-3)(12j-1)`

`=72(j^2)-36j-6j+3`

`=-69-42j`

**f. **`(3 + 2j)(3 − 2j)`

Answer

`(3 + 2j)(3 − 2j) `

`= (3)^2 − (2j)^2`

`= 9 − 4j^2`

`= 9 + 4`

`= 13`

Please support IntMath!

### Multiplying by the conjugate

Example 2(f) is a special case.

`3 + 2j` is the **conjugate** of `3 − 2j`.

In general:

`x + yj` is the

conjugateof `x − yj`and

`x − yj` is the

conjugateof `x + yj`.

Notice that when we multiply conjugates, our final answer is **real** only (it does not contain any imaginary terms.

We use the idea of **conjugate** when dividing complex numbers**.**

## Division of Complex Numbers

Earlier, we learned how to rationalise the denominator of an expression like:

`5/(3-sqrt2)`

To simplify the expression, we multiplied numerator and denominator by the **conjugate** of the denominator, `3 + sqrt2` as follows:

`5/(3-sqrt2)xx(3+sqrt2)/(3+sqrt2)`

`=(15+5sqrt2)/(9-2)`

`=(15+5sqrt2)/7`

We did this so that we would be left with no radical (square root) in the denominator.

Dividing by a complex number is a similar process to the above - we multiply top and bottom of the fraction by the conjugate of the bottom.

### Example 3 - Division

**a**. Express

`(3-j)/(4-2j)`

in the form *x *+* yj.*

Answer

The **conjugate** of `4 − 2j` is `4 +
2j`. We multiply the top and bottom of the fraction by this conjugate.

`(3-j)/(4-2j) xx (4+2j)/(4+2j)`

`=(12+6j-4j-2j^2)/(16-4j^2)`

`=(12+2+6j-4j)/(16+4)`

`=(14+2j)/20`

`=(7+j)/10`

Please support IntMath!

**b**. Simplify:

`(1-sqrt(-4))/(2+9j)`

Answer

We multiply the top and bottom of the fraction by the conjugate of the bottom (denominator).

`(1-sqrt(-4))/(2+9j)`

` =(1-2j)/(2+9j) xx (2-9j)/(2-9j)`

`=(2-9j-4j+18j^2)/(4-81j^2)`

`=(-16-13j)/(4+81)`

`=(-16-13j)/85`

Please support IntMath!

**Exercises**

**1. **Express in the form *a *+* bj*:

`(4+sqrt(-16))+(3-sqrt(-81))`

Answer

`(4+sqrt(-16))+(3-sqrt(-81))`

`=(4+4j)+(3-9j)`

`=7-5j`

Please support IntMath!

**2. **Express in the form *a *+* bj.*

`sqrt(-4)/(2+sqrt(-9))`

Answer

We multiply the top and bottom of the fraction by the conjugate of the bottom (denominator).

`(2j)/(2+3j) xx (2-3j)/(2-3j)`

` = (4j-6j^2)/(4+9)`

`=(6+4j)/13`

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!