# 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)`

**b. **`(6 − j)(5j)`

**c. **`(2 − j)(3 + j)`

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

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

**f. **`(3 + 2j)(3 − 2j)`

### 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)`

We multiplied numerator and denominator by the **conjugate** of the denominator, `3 + sqrt2`:

`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 with complex numbers is similar.

### Example 3 - Division

**a**. Express

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

in the form *x *+* yj.*

**b**. Simplify:

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

**Exercises**

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

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

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

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

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