1.3 Complex Numbers Basic Concepts of Complex Numbers Operations on Complex Numbers
1.3 - 1 Basic Concepts of Complex Numbers
There are no real numbers for the solution of the equation x 2 = −1. To extend the real number system to include such numbers as, −1, the number i is defined to have the following property; i 2 = −1.
1.3 - 2 Basic Concepts of Complex Numbers So… i = −1 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.
1.3 - 3 Nonreal complex Complex numbers numbers a + bi, a + bi, b ≠ 0 Irrational a and b numbers real Real numbers Integers a + bi, b = 0 Rational numbers Non- integers Basic Concepts of Complex Numbers
Two complex numbers are equal provided that their real parts are equal and their imaginary parts are equal; a+=+ bi c di if and only if ac= and bd=
1.3 - 5 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.
1.3 - 6 THE EXPRESSION −a
If a > 0, then −=a ia.
1.3 - 7 Example 1 WRITING − a AS ia Write as the product of a real number and i, using the definition of −a. a. −16 Solution:
−16= i 16= 4i
1.3 - 8 Example 1 WRITING − a AS ia Write as the product of a real number and i, using the definition of −a. b. −70 Solution:
−=70i 70
1.3 - 9 Example 1 WRITING − a AS ia Write as the product of a real number and i, using the definition of −a. c. − 48
Solution: Product rule −=48ii 48 = 16 3 = 4 i 3 for radicals
1.3 - 10 Operations on Complex Numbers
Products or quotients with negative radicands are simplified by first rewriting −a as iafor a positive number.
Then the properties of real numbers are applied, together with the fact that i 2 = −1.
1.3 - 11 FINDING PRODUCTS AND Example 2 QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer. a. −−77
Solution: −7 −= 7ii 77 2 First write all square 2 roots in terms of i. = i ( 7 ) = −17 i 2 = −1 = −7
1.3 - 12 FINDING PRODUCTS AND Example 2 QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer. b. −−6 10
Solution: −6 −= 10ii 6 10 = −1 2 15
= i 2 60 = −2 15 = −1 415
1.3 - 13 FINDING PRODUCTS AND Example 2 QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer. −20 c. −2 Solution: −20i 20 20 Quotient = = = 10 rule for −22i 2 radicals
1.3 - 14 Example 3 SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND Write −+8 − 128 in standard form a + bi. 4 Solution: −+8 −128 −+8 −64 2 = 44
−+8 8i 2 = −=64 8i 4
1.3 - 15 Example 3 SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND Write −+8 − 128 in standard form a + bi. 4 Solution: −+8 8i 2 = −=64 8i 4 Be sure to factor before 4(−+22i 2) simplifying = Factor. 4 =−+222i Lowest terms
1.3 - 16 Addition and Subtraction of Complex Numbers For complex numbers a + bi and c + di, (a+ bi )( ++ c di )( =+++ a c )( b d ) i and (a+ bi ) −+ ( c di ) =−+− ( a c ) ( b d ). i
1.3 - 17 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference. a. (3− 4)ii +− ( 2 + 6) Add Solution: Add real imaginary parts. parts. (3−−4ii )(+− 23 +=66 )[ ++(−2)] [ 4 +]i Commutative, associative, distributive properties =12 + i
1.3 - 18 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference. b. (−+ 9 7)ii + (3 − 15) Solution:
(−+ 9 7)iii + (3 − 15) =−− 6 8
1.3 - 19 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference. c. (−+ 4 3)ii − (6 − 7) Solution:
(−+−− 4 3ii ) (6 7 ) =−−+−− ( 4 6)[ 3 ( 7)]i
=−+10 10i
1.3 - 20 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference. d. (12− 5)ii −− (8 3) Solution:
(12− 5)i −− (8 3) ii =− 4 2
1.3 - 21 Multiplication of Complex Numbers For complex numbers a + bi and c + di,
(a+ bi )( c +=−++ di ) ( ac bd ) ( ad bc ) i .
1.3 - 22 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product. a. (2−+ 3)(3ii 4) Solution: (2− 3)(3i += 4) i 2(3) + 2(4) i − 3(3) i − 3(4) iiFOIL =+−−6 8ii 9 12i 2
=6 −−i 12(−1 ) i2 = −1 =18 − i
1.3 - 23 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product. b. (4+ 3i )2 Solution: (4+=+ 3)i22 4 2(4)(3) ii + (3)2Square of a binomial Remember to add twice the =++16 24ii 9 2 product of the two terms.
=16 + 24i +− 9( 1) i 2 = −1
=7 + 24i
1.3 - 24 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product. c. (6+− 5ii )(6 5 )
Solution: Product of the sum (6+ 5)(6ii −=− 5) 622 (5) i and difference of two terms
=−−36 25( 1) i 2 = −1 =36 + 25 =61, or 61 + 0i Standard form
1.3 - 25 Simplifying Powers of i
Powers of i can be simplified using the facts
ii2=−===1 and 42 (i 2 ) (−1) 2 1
1.3 - 26 Example 6 SIMPLIFYING POWERS OF i
Simplify each power of i. a. i 15 Solution: Since i 2 = –1 and i 4 = 1, write the given power as a product involving i 2 or i 4. For example, i3 =i 2 i =()−1 ii = − . Alternatively, using i4 and i3 to rewrite i15 gives 3 i15= ii 12 3 =(i 4 ) i33 =1 ( −=− i) i
1.3 - 27 Powers of i
ii1 = ii5 = ii9 = i 2 = −1 i 6 = −1 i 10 = −1 ii3 = − ii7 = − ii11 = − i 4 = 1 i 8 = 1 i 12 = 1, and so on.
1.3 - 28 Ex 5c. showed that…
(6+ 5ii )(6 −= 5 ) 61
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.
1.3 - 29 Property of Complex Conjugates For real numbers a and b,
(a+ bi )( a −=+ bi ) a22 b .
1.3 - 30 Example 7 DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi. a. 32+ i 5 − i Solution: Multiply by the complex conjugate of 32+ ii (32)+ (5 + i) = the denominator in 5(− i 5)− i ()5 + i both the numerator and the denominator. 15++ 3i 10 ii + 2 2 = Multiply. 25 − i 2
1.3 - 31 Example 7 DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi. a. 32+ i 5 − i Solution: 15++ 3i 10 ii + 2 2 = Multiply. 25 − i 2 13+ 13i = i 2 = −1 26
1.3 - 32 Example 7 DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi. a. 32+ i 5 − i Solution: 13+ 13i = i 2 = −1 26 a+ bi a bi 13 13i = + = + c cc 26 26
1.3 - 33 Example 7 DIVIDING COMPLEX NUMBERS
Write each quotient in standard form a + bi. a. 32+ i 5 − i Solution: a+ bi a bi 13 13i = + = + c cc 26 26 11 = + i Lowest terms; 22 standard form
1.3 - 34