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Complex Numbers Objectives Students will learn:

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Presentation on theme: "Complex Numbers Objectives Students will learn:"— Presentation transcript:

1 Complex Numbers Objectives Students will learn:
Basic Concepts of Complex Numbers Operations on Complex Numbers

2 Basic Concepts of Complex Numbers
There are no real numbers for the solution of the equation To extend the real number system to include such numbers as, the number i is defined to have the following property;

3 Basic Concepts of Complex Numbers
The number i is called the imaginary unit. Numbers of the form a + bi, where a and b are real numbers are called complex numbers. In this complex number, a is the real part and b is the imaginary part.

4 Basic Concepts of Complex Numbers
Two complex numbers are equal provided that their real parts are equal and their imaginary parts are equal; if and only if and

5 Basic Concepts of Complex Numbers
For complex number a + bi, if b = 0, then a + bi = a So, the set of real numbers is a subset of complex numbers.

6 Nonreal complex numbers
a + bi, b ≠ 0 Complex numbers a + bi, a and b real Irrational numbers Real numbers a + bi, b = 0 Integers Rational numbers Non-integers

7 Basic Concepts of Complex Numbers
If a = 0 and b ≠ 0, the complex number is pure imaginary. A pure imaginary number or a number, like 7 + 2i with a ≠ 0 and b ≠ 0, is a nonreal complex number. The form a + bi (or a + ib) is called standard form.

8 THE EXPRESSION

9 Write as the product of a real number and i, using the definition of
Example 1 WRITING AS Write as the product of a real number and i, using the definition of a. Solution:

10 Write as the product of a real number and i, using the definition of
Example 1 WRITING AS Write as the product of a real number and i, using the definition of b. Solution:

11 Write as the product of a real number and i, using the definition of
Example 1 WRITING AS Write as the product of a real number and i, using the definition of c. Solution: Product rule for radicals

12 Operations on Complex Numbers
Products or quotients with negative radicands are simplified by first rewriting for a positive number. Then the properties of real numbers are applied, together with the fact that

13 Operations on Complex Numbers
Caution When working with negative radicands, use the definition… before using any of the other rules for radicands.

14 Operations on Complex Numbers
Caution In particular, the rule is valid only when c and d are not both negative. while so

15 First write all square roots in terms of i.
FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS Example 2 Multiply or divide, as indicated. Simplify each answer. a. Solution: First write all square roots in terms of i. i 2 = −1

16 Multiply or divide, as indicated. Simplify each answer.
FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS Example 2 Multiply or divide, as indicated. Simplify each answer. c. Solution: Quotient rule for radicals

17 Addition and Subtraction of Complex Numbers
For complex numbers a + bi and c + di, and

18 Find each sum or difference.
ADDING AND SUBTRACTING COMPLEX NUMBERS Example 4 Find each sum or difference. a. Add imaginary parts. Solution: Add real parts. Commutative, associative, distributive properties

19 Find each sum or difference.
ADDING AND SUBTRACTING COMPLEX NUMBERS Example 4 Find each sum or difference. b. Solution:

20 Find each sum or difference.
ADDING AND SUBTRACTING COMPLEX NUMBERS Example 4 Find each sum or difference. c. Solution:

21 Find each sum or difference.
ADDING AND SUBTRACTING COMPLEX NUMBERS Example 4 Find each sum or difference. d. Solution:

22 Multiplication of Complex Numbers
The product of two complex numbers is found by multiplying as if the numbers were binomials and using the fact that i2 = – 1, as follows. FOIL Distributive property; i 2 = – 1

23 Multiplication of Complex Numbers
For complex numbers a + bi and c + di,

24 Find each product. a. Solution: MULTIPLYING COMPLEX NUMBERS Example 5
FOIL i2 = −1

25 Remember to add twice the product of the two terms.
MULTIPLYING COMPLEX NUMBERS Example 5 Find each product. b. Solution: Square of a binomial Remember to add twice the product of the two terms. i 2 = −1

26 Find each product. c. Solution: MULTIPLYING COMPLEX NUMBERS Example 5
Product of the sum and difference of two terms i 2 = −1 Standard form

27 Simplifying Powers of i
Powers of i can be simplified using the facts

28 Simplify each power of i.
SIMPLIFYING POWERS OF i Example 6 Simplify each power of i. a. Solution: Since i 2 = – 1 and i 4 = 1, write the given power as a product involving i 2 or i 4. For example, Alternatively, using i4 and i3 to rewrite i15 gives

29 Simplify each power of i.
SIMPLIFYING POWERS OF i Example 6 Simplify each power of i. b. Solution:

30 Powers of i and so on.

31 Ex 5c. showed that… The numbers differ only in the sign of their imaginary parts and are called complex conjugates. The product of a complex number and its conjugate is always a real number. This product is the sum of squares of real and imaginary parts.

32 Property of Complex Conjugates
For real numbers a and b,

33 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. a. Solution: Multiply by the complex conjugate of the denominator in both the numerator and the denominator. Multiply.

34 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. a. Solution: Multiply. i 2 = −1

35 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. a. Solution: i 2 = −1

36 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. a. Solution: Lowest terms; standard form

37 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. b. Solution: – i is the conjugate of i.

38 Write each quotient in standard form a + bi.
DIVIDING COMPLEX NUMBERS Example 7 Write each quotient in standard form a + bi. b. Solution: Standard form i 2 = −1(−1) = 1


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