# 16.4.1: Complex Numbers

- Page ID
- 73010

- Express roots of negative numbers in terms of \(\ i\).
- Express imaginary numbers as \(\ bi\) and complex numbers as \(\ a+b i\).

## Introduction

Several times in your learning of mathematics, you have been introduced to new kinds of numbers. Each time, these numbers made possible something that seemed impossible! Before you learned about negative numbers, you couldn’t subtract a greater number from a lesser one, but negative numbers give us a way to do it. When you were learning to divide, you initially weren't able to do a problem like 13 divided by 5 because 13 isn't a multiple of 5. You then learned how to do this problem writing the answer as 2 remainder 3. Eventually, you were able to express this answer as \(\ 2 \frac{3}{5}\). Using fractions allowed you to make sense of this division.

Up to now, you’ve known it was impossible to take a square root of a negative number. This is true, using only the *real numbers*. But here you will learn about a new kind of number that lets you work with square roots of negative numbers! Like fractions and negative numbers, this new kind of number will let you do what was previously impossible.

## Using \(\ i\) to Simplify Roots of Negative Numbers

You really need only one new number to start working with the square roots of negative numbers. That number is the square root of -1, \(\ \sqrt{-1}\). The *real numbers* are those that can be shown on a number line; they seem pretty *real* to us! When something’s not real, you often say it is *imaginary*. So let’s call this new number \(\ i\) and use it to represent the square root of -1.

\[\ i=\sqrt{-1} \nonumber \]

Because \(\ \sqrt{x} \cdot \sqrt{x}=x\), we can also see that \(\ \sqrt{-1} \cdot \sqrt{-1}=-1\) or \(\ i \cdot i=-1\). We also know that \(\ i \cdot i=i^{2}\), so we can conclude that \(\ i^{2}=-1\).

\[\ i^{2}=-1 \nonumber \]

The number \(\ i\) allows us to work with roots of all negative numbers, not just \(\ \sqrt{-1}\). There are two important rules to remember: \(\ \sqrt{-1}=i\), and \(\ \sqrt{a b}=\sqrt{a} \sqrt{b}\). You will use these rules to rewrite the square root of a negative number as the square root of a positive number times \(\ \sqrt{-1}\). Next you will simplify the square root and rewrite \(\ \sqrt{-1}\) as \(\ i\). Let’s try an example.

**Simplify. \(\ \sqrt{-4}\)**

**Solution**

\(\ \sqrt{-4}=\sqrt{(4)(-1)}=\sqrt{4} \sqrt{-1}\) | Use the rule \(\ \sqrt{a b}=\sqrt{a} \sqrt{b}\) to rewrite this as a product using \(\ \sqrt{-1}\). |

\(\ \sqrt{4} \sqrt{-1}=2 \sqrt{-1}\) | Since 4 is a perfect square \(\ \left(4=2^{2}\right)\), you can simplify the square root of 4. |

\(\ 2 \sqrt{-1}=2 i\) | Use the definition of \(\ i\) to rewrite \(\ \sqrt{-1}\) as \(\ i\). |

\(\ \sqrt{-4}=2 i\)

**Simplify. \(\ \sqrt{-18}\)**

**Solution**

\(\ \sqrt{-18}=\sqrt{18 \cdot-1}=\sqrt{18} \sqrt{-1}\) | Use the rule \(\ \sqrt{a b}=\sqrt{a} \sqrt{b}\) to rewrite this as a product using \(\ \sqrt{-1}\). |

\(\ \sqrt{18} \sqrt{-1}=\sqrt{9} \sqrt{2} \sqrt{-1}=3 \sqrt{2} \sqrt{-1}\) | Since 18 is not a perfect square, use the same rule to rewrite it using factors that are perfect squares. In this case, 9 is the only perfect square factor, and the square root of 9 is 3. |

\(\ 3 \sqrt{2} \sqrt{-1}=3 \sqrt{2 i}=3 i \sqrt{2}\) |
Use the definition of \(\ i\) to rewrite \(\ \sqrt{-1}\) as \(\ i\). Remember to write \(\ i\) in front of the radical. |

\(\ \sqrt{-18}=3 i \sqrt{2}\)

**Simplify. \(\ -\sqrt{-72}\)**

**Solution**

\(\ -\sqrt{-72}=-\sqrt{72 \cdot-1}=-\sqrt{72} \sqrt{-1}\) | Use the rule \(\ \sqrt{a b}=\sqrt{a} \sqrt{b}\) to rewrite this as a product using \(\ \sqrt{-1}\) |

\(\ -\sqrt{72} \sqrt{-1}=-\sqrt{36} \sqrt{2} \sqrt{-1}=-6 \sqrt{2} \sqrt{-1}\) |
Since 72 is Notice that 72 has three perfect squares as factors: 4, 9, and 36. It’s easiest to use the largest factor that is a perfect square. |

\(\ -6 \sqrt{2} \sqrt{-1}=-6 \sqrt{2} i=-6 i \sqrt{2}\) |
Use the definition of \(\ i\) to rewrite \(\ \sqrt{-1}\) as \(\ i\). Remember to write \(\ i\) in front of the radical. |

\(\ -\sqrt{-72}=-6 i \sqrt{2}\)

You may have wanted to simplify \(\ -\sqrt{-72}\) using different factors. Some may have thought of rewriting this radical as \(\ -\sqrt{-9} \sqrt{8}\), or \(\ -\sqrt{-4} \sqrt{18}\), or \(\ -\sqrt{-6} \sqrt{12}\) for instance. Each of these radicals would have eventually yielded the same answer of \(\ -6 i \sqrt{2}\).

- Find perfect squares within the radical.
- Rewrite the radical using the rule \(\ \sqrt{a b}=\sqrt{a} \cdot \sqrt{b}\).
- Rewrite \(\ \sqrt{-1}\) as \(\ i\).

Example: \(\ \sqrt{-18}=\sqrt{9} \sqrt{-2}=\sqrt{9} \sqrt{2} \sqrt{-1}=3 i \sqrt{2}\)

Simplify. \(\ \sqrt{-50}\)

- \(\ 5\)
- \(\ -5 \sqrt{2}\)
- \(\ 5 i\)
- \(\ 5 i \sqrt{2}\)

**Answer**-
- \(\ 5\)
Incorrect. You may have noticed the perfect square 25 as a factor of 50, but forgot the rest of the number under the radical. The correct answer is: \(\ \sqrt{-50}=\sqrt{25} \sqrt{2} \sqrt{-1}=5 \sqrt{2 i}=5 i \sqrt{2}\).

- \(\ -5 \sqrt{2}\)
Incorrect. While \(\ \sqrt{50}=5 \sqrt{2}\), the negative under the radical cannot be moved outside the radical. Remember, \(\ \sqrt{-1}=i\). The correct answer is: \(\ \sqrt{-50}=\sqrt{25} \sqrt{2} \sqrt{-1}=5 \sqrt{2 i}=5 i \sqrt{2}\).

- \(\ 5 i\)
Incorrect. You may have correctly noticed the perfect square 25 as a factor of 50, and correctly used \(\ \sqrt{-1}=i\), but you forgot the remaining factor of -50, which is 2. The correct answer is: \(\ \sqrt{-50}=\sqrt{25} \sqrt{2} \sqrt{-1}=5 \sqrt{2 i}=5 i \sqrt{2}\).

- \(\ 5 i \sqrt{2}\)
Correct. \(\ \sqrt{-50}=\sqrt{25} \sqrt{2} \sqrt{-1}=5 \sqrt{2 i}=5 i \sqrt{2}\).

- \(\ 5\)

## Imaginary and Complex Numbers

You can create other numbers by multiplying \(\ i\) by a real number. An **imaginary number** is any number of the form \(\ bi\), where \(\ b\) is real (but not 0) and \(\ i\) is the square root of -1. Look at the following examples, and notice that \(\ b\) can be any kind of real number (positive, negative, whole number, rational, or irrational), but not 0. (If \(\ b\) is 0, \(\ 0i\) would just be 0, a real number.)

\(\ \begin{aligned} \text { 3i } b&=3 \\ \frac{3}{5} \text { i } b&=\frac{3}{5} \end{aligned}\) |
\(\ \begin{array}{rl} -672 i & b=-672 \\ \frac{\sqrt{2}}{2} i & b=\frac{\sqrt{2}}{2} \end{array}\) |

You can use the usual operations (addition, subtraction, multiplication, and so on) with imaginary numbers. You’ll see more of that, later. When you *add* a real number to an imaginary number, however, you get a **complex number**. A complex number is any number in the form \(\ a+b i\), where \(\ a\) is a real number and \(\ bi\) is an imaginary number. The number \(\ a\) is sometimes called the **real part** of the complex number, and \(\ bi\) is sometimes called the **imaginary part**.

Complex Number |
Real part |
Imaginary part |
---|---|---|

\(\ 3+7 i\) | \(\ 3\) | \(\ 7 i\) |

\(\ 18-32 i\) | \(\ 18\) | \(\ -32 i\) |

\(\ -\frac{3}{5}+i \sqrt{2}\) | \(\ -\frac{3}{5}\) | \(\ i \sqrt{2}\) |

\(\ \frac{\sqrt{2}}{2}-\frac{1}{2} i\) | \(\ \frac{\sqrt{2}}{2}\) | \(\ -\frac{1}{2} i\) |

In a number with a radical as part of \(\ b\), such as \(\ -\frac{3}{5}+i \sqrt{2}\) above, the imaginary \(\ i\) should be written in front of the radical. Though writing this number as \(\ -\frac{3}{5}+\sqrt{2} i\) is technically correct, it makes it much more difficult to tell whether \(\ i\) is inside or outside of the radical. Putting it before the radical, as in \(\ -\frac{3}{5}+i \sqrt{2}\), clears up any confusion. Look at these last two examples.

Number |
Number in complex form: \(\ a+b i\) |
Real part |
Imaginary part |
---|---|---|---|

\(\ 17\) | \(\ 17+0 i\) | \(\ 17\) | \(\ 0 i\) |

\(\ -3 i\) | \(\ 0-3 i\) | \(\ 0\) | \(\ -3 i\) |

By making \(\ b=0\), any real number can be expressed as a complex number. The real number \(\ a\) is written \(\ a+0 i\) in complex form. Similarly, any imaginary number can be expressed as a complex number. By making \(\ a=0\), any imaginary number \(\ bi\) is written \(\ 0+b i\) in complex form.

**Write 83.6 as a complex number. **

**Solution**

\(\ a+b i\) | Remember that a complex number has the form \(\ a+b i\). You need to figure out what \(\ a\) and \(\ b\) need to be. |

\(\ 83.6+b i\) | Since 83.6 is a real number, it is the real part \(\ (a)\) of the complex number \(\ a+b i\). A real number does not contain any imaginary parts, so the value of \(\ b\) is 0. |

\(\ 83.6+0 i\)

**Write \(\ -3 i\) as a complex number.**

**Solution**

\(\ a+b i\) | Remember that a complex number has the form \(\ a+b i\). You need to figure out what \(\ a\) and \(\ b\) need to be. |

\(\ a-3 i\) | Since \(\ -3 i\) is an imaginary number, it is the imaginary part \(\ (bi)\) of the complex number \(\ a+b i\). This imaginary number has no real parts, so the value of \(\ a\) is 0. |

\(\ 0-3 i\)

Which is the real part of the complex number \(\ -35+9 i\)?

- 9
- -35
- 35
- 9 and -35

**Answer**-
- 9
Incorrect. The number 9 is in the imaginary part \(\ (9i)\) of this complex number. In a complex number, \(\ a+b i\), \(\ a\) is the real part. In this case, \(\ a=-35\), so the real part is -35.

- -35
Correct. In a complex number, \(\ a+b i\), \(\ a\) is the real part. In this case, \(\ a=-35\), so the real part is -35.

- 35
Incorrect. In a complex number, \(\ a+b i\), \(\ a\) is the real part. In this case, \(\ a=-35\), The real part can be any real number, including negative numbers.

- 9 and -35
Incorrect. The number 9 is in the imaginary part \(\ (9i)\) of this complex number. In a complex number, \(\ a+b i\), \(\ a\) is the real part. In this case, \(\ a=-35\), so the real part is just -35.

- 9

## Summary

Complex numbers have the form \(\ a+b i\), where \(\ a\) and \(\ b\) are real numbers and \(\ i\) is the square root of -1. All real numbers can be written as complex numbers by setting \(\ b=0\). Imaginary numbers have the form \(\ bi\) and can also be written as complex numbers by setting \(\ a=0\). Square roots of negative numbers can be simplified using \(\ \sqrt{-1}=i\) and \(\ \sqrt{a b}=\sqrt{a} \sqrt{b}\).