Section 1.6

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Section 1.6

The Complex Number System






Topics

o The imaginary unit i and its propertieso The algebra of complex numberso Roots and complex numbers






The Imaginary Unit i and its Properties

The Imaginary Unit iThe imaginary unit i is defined as In other words, i has the property that its square is −1:

.1i2 .1i







Square Roots of Negative NumbersIf a is a positive real number, Note that by this definition, and by a logical extension of exponentiation,

.a i a

2 22 .i a i a a






Example 1

As is customary, we write a constant such as 4 before letters in algebraic expressions, even if, as in this case, the letter is not a variable. Remember that i has a fixed meaning: i is the square root of −1.

As is customary, again, we write the radical factor last. You should verify that is indeed −8.

16a.

8b.

22 2i

16i 4i .4i

8i 2 2i .2 2i






Example 1 (cont.)

The simple fact that i2 = −1 allows us, by our extension of exponentiation, to determine in for any natural number n.

This observation shows that −i also has the property that its square is −1.

3ic.

2id.

2 1i i i , i and 4i 2 2 1 1i i .1

2 2 21 i i 1.







Complex NumbersFor any two real numbers a and b, the sum a + bi is a complex number. The collection = {a + bi|a and b are both real} is called the set of complex numbers and is another example of a field. The number a is called the real part of a + bi, and the number b is called the imaginary part. If the imaginary part of a given complex number is 0, the number is simply a real number. If the real part of a given complex number is 0, the number is a pure imaginary number.






The Algebra of Complex Numbers

Simplifying Complex Expressions Step 1: Add, subtract, or multiply the complex

numbers, as required, by treating every complex number a + bi as a polynomial expression. Remember, though, that i is not actually a variable. Treating a + bi as a binomial in i is just a handy device.

Step 2: Complete the simplification by using the fact that i2 = −1.






Example 2

Simplify the following complex number expressions.

Solutions:

2

4 3 5 7 2 3 3. . 3

3 2 2 3 2 3. .

i i i i

i i i

a b

c d

5. 4 3 7i i a Treating the two complex numbers as polynomials in i, we combine the real parts and then the imaginary parts.

4 5 3 7 i

1 10i






Example 2 (cont.)

We begin by distributing the minus sign over the two terms of the second complex number, and then combine as in part a.

2 3 3 3i i b.

2 3 3 3i i

2 3 3 3 i

1 0i 1






Example 2 (cont.)

The product of two complex numbers leads to four products via the distributive property, as illustrated here. After multiplying, we combine the two terms containing i, and rewrite i2 as −1.

3 2 2 3i i c.26 9 4 6i i i

6 9 4 6 1i

6 5 6i 12 5i






Example 2 (cont.)

Squaring this complex number also leads to four products, which we simplify as in part c. Remember that a complex number is not simplified until it has the form a + bi.

22 3id. 2 3 2 3i i 24 6 6 9i i i

4 12 9 1i

5 12i






Example 3

Simplify the quotients.

1

2 33

4 3

1

ii

i

i

a.

b.

c.






Example 3(cont.)

Solutions:2 33

ii

a. We multiply the top and bottom of the fraction by 3 + i, which is the complex conjugate of the denominator. The rest of the simplification involves multiplying complex numbers as in the last example.

We would often leave the answer

in the form unless it is necessary to identify the real and imaginary parts.

3 1110

i

2 3 33 3

i ii i

2 3 33 3

i ii i

2

2

6 2 9 39 3 3

i i ii i i

3 11

10i

3 11

10 10i






Example 3(cont.)

14 3i b. In this example, we simplify the reciprocal of the complex number 4 − 3i. After writing the original expression as a fraction, we multiply the top and bottom by the complex conjugate of the denominator and proceed as in part a.

14 3i

1 4 34 3 4 3

ii i

4 3

4 3 4 3i

i i

2

4 316 9

ii

4 3

25i

4 3

25 25i






Example 3(cont.)

1i

c. This problem illustrates the process of writing the reciprocal of the imaginary unit as a complex number. Note that with this as a starting point, we could now calculate i –2, i –3, …

1 ii i

2

ii

1i

i






Example 4

Simplify the following expressions.

22 3

44

a.

b.






Example 4 (cont.)

Solutions:

22 3 a. Each is converted

to before carrying out the associated multiplications. Note that incorrect use of one of the properties of radicals would have led to adding 3 instead of subtracting 3.

33i 2 3 2 3

4 4 3 3 3

24 4 3 3i i

4 4 3 3i

1 4 3i






Example 4 (cont.)

44

b. 22i

1i

i

We have already simplified in Example 3c, so we quickly obtain the correct answer of −i. If we had incorrectly rewritten the original fraction as

we would have obtained or i as the

final answer.

1i

4,

41

Section 1.6

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Transcript of Section 1.6