# What does 12 6 6 equal to

30

## COMPLEX

OR IMAGINARY

NUMBERS

The defining property of *i*

The square root of a negative number

Powers of *i*

Algebra with complex numbers

The real and imaginary components

Complex conjugates

IN ALGEBRA, we want to be able to say that every polynomial equation has a solution; specifically, this one:

*x*^{2} + 1 = 0.

That implies,

*x*^{2} = −1.

But there is no real number whose square is negative. (Lesson 13.) Therefore, we invent a number—a complex or imaginary number—and we call it *i*.

*i*^{2} = −1.

That is the defining property of the complex unit*i*. Its *square* is negative.

In other words,

*i* = .

The complex number *i* is purely algebraic. That is, we call it a "number" because it will obey all the rules we normally associate with a number. We may add it, subtract it, multiply it, and so on. The complex number *i* turns out to be extremely useful in mathematics and physics.

Example 1. 3*i* **·** 4*i*= 12*i*^{2} = 12(−1) = −12.

Example 2. −5*i* **·** 6*i* = −30*i*^{2} = 30.

We see, then, that the factor *i*^{2} changes the sign of a product.

Problem 1. Evaluate the following.

To see the answer, pass your mouse over the colored area.

To cover the answer again, click "Refresh" ("Reload").

Do the problem yourself first!

a) | i^{2} = −1 | b) | i· 2i = 2i^{2} = 2(−1) = −2 | |

c) | (3i)^{2} = 3^{2}i^{2} = −9 | d) | −5i· 4i = −20i^{2} = 20 |

The square root of a negative number

If a radicand is negative—

, where *a* > 0,

—then we can simplify it as follows:

= = = *i*.

In other words:

The square root of −*a* is equal to *i* times the square root of *a*.

Problem 2. Express each of the following in terms of *i*.

Powers of *i*

Let us begin with *i*^{0}, which is 1. (Any number with exponent 0 is 1.) Each power of *i* can be obtained from the previous power by multiplying it by *i*. We have:

i^{0} | = | 1 |

i^{1} | = | i |

i^{2} | = | −1 |

i^{3} | = | −1· i = −i |

i^{4} | = | −i· i = −i^{2} = −(−1) = 1 |

And we are back at 1—the cycle of powers will repeat. Any power of *i* will be either

1, *i*, −1, or −*i*

—according to the remainder upon dividing the exponent *n* by 4.

Examples 4.

i^{9} | = | i, because on dividing 9 by 4, the remainder is 1. | |

i^{9} = i^{1}. | |||

i^{18} | = | −1, because on dividing 18 by 4, the remainder is 2. | |

i^{18} = i^{2}. | |||

i^{35} | = | −i, because on dividing 35 by 4, the remainder is 3. | |

i^{35} = i^{3}. | |||

i^{40} | = | 1, because on dividing 40 by 4, the remainder is 0. | |

i^{40} = i^{0}. |

**Note:** Even powers of *i* will be either 1 or −1, according as the exponent is a multiple of 4, or 2 more than a multiple of 4. While odd powers will be either *i* or −*i*.

Problem 3. Evaluate each power of *i*.

a) i^{3} = −i | b) i^{4} = 1 | c) i^{6} = i^{2} = −1 | ||

d) i^{9} = i^{1} = i | e) i^{12} = i^{0} = 1 | f) i^{17} = i^{1} = i | ||

g) i^{27} = i^{3} = −i | h) i^{30} = i^{2} = −1 | i) i^{100} = i^{0} = 1 |

Problem 4. |

Algebra with complex numbers

Complex numbers follow the same rules as real numbers. For example, to multiply

(2 + 3*i*)(2 − 3*i*)

the student should recognize the form (*a* + *b*)(*a* − *b*) -- which will produce the difference of two squares. Therefore,

(2 + 3i)(2 − 3i) | = | 4 − 9i^{2} |

= | 4 − 9(−1) | |

= | 4 + 9 | |

= | 13. |

Again, the factor *i*^{2}changes the sign of the term.

Problem 5. Multiply.

a) (1 + *i*)(1 − *i*) = 1 − 2*i*^{2} = 1 + 2 = 3

c) (2 + 3i)(4 − 5i) | = | 8 − 10i + 12i − 15i^{2} |

= | 8 + 2i + 15 | |

= | 23 + 2i |

Problem 6. (*x* + 1 + 3*i*)(*x* + 1 − 3*i*)

a) What form will that produce? The difference of two squares.

Lesson 19.

b) Multiply out.

(x + 1 + 3i)(x + 1 − 3i) | = | (x + 1)^{2} − 9i^{2} |

= | x^{2} + 2x + 1 + 9 | |

= | x^{2} + 2x + 10 |

c) | (x − 2 − i)(x − 2 + i) | = | (x − 2)^{2} − 2i^{2} |

= | x^{2} − 4x + 4 + 2 | ||

= | x^{2} − 4x + 6 |

The real and imaginary components

Here is the standard form of a complex number:

*a* + *bi*.

Both *a* and *b* are real. For example,

3 + 2*i*.

*a*—that is, 3 in the example—is called the real component (or the real part). *b* (2 in the example) is called the imaginary component (or the imaginary part).

Again, the components are real.

Problem 7. Name the real component *a* and the imaginary component *b*.

a) | 3 − 5i a = 3, b = −5. | b) | 1 + i a = 1, b = . | |

c) | i a = 0, b = 1. | d) | −6 a = −6, b = 0. |

Complex conjugates

The complex conjugate of *a* + *bi* is *a* − *bi*. The main point about a conjugate pair is that when they are multiplied—

(*a* + *bi*)(*a* − *bi*)

—the product is a positive real number. For, that form is the difference of two squares:

(*a* + *bi*)(*a* − *bi*) = *a*^{2} − *b*^{2}*i*^{2} = *a*^{2} + *b*^{2}

The product of a complex conjugate pair

is equal to the __sum__ of the squares of the components.

Problem 8. Calculate the positive real number that results from multiplying each number with its complex conjugate.

a) 2 + 3*i*

(2 + 3*i*)(2 − 3*i*) = 2^{2} + 3^{2} = 4 + 9 = 13.

b) 3 − *i*.

(3 − *i*)(3 + *i*) = = 3^{2} + ()^{2}= 9 + 2 = 11.

c) *u* + *iv*. (*u* + *iv*)(*u* − *iv*) = *u*^{2} + *v*^{2}.

d) 1 + *i*. (1 + *i*)(1 − *i*) = 1^{2} + 1^{2} = 2.

e) −*i*. (−*i*)(*i*) = −*i*^{2} = 1.

Next Lesson: Rectangular coördinates

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