Chapter 7 Radicals, Radical Functions, and Rational Exponents.

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Presentation transcript:

Chapter 7 Radicals, Radical Functions, and Rational Exponents

7.1 Radical Expressions and Functions Square Root Square Root If a>= 0, then b >= 0, such that b 2 = a, is the principal square root of a If a>= 0, then b >= 0, such that b 2 = a, is the principal square root of a √ a = b √ a = b E.g., E.g., √25 = 5 √25 = 5 √100 = 10 √100 = 10

= ----, because --- = = ----, because --- = = 25 = = 25 = = = = = 7

Negative Square Root 25 = principal square root 25 = principal square root - 25 = negative square root - 25 = negative square root Given: a Given: a What is the square root of a? What is the square root of a? Given: 25 Given: 25 What is the square root of 25? What is the square root of 25? sqrt = 5, sqrt = -5, because 5 2 = 25, (-5) 2 = 25 sqrt = 5, sqrt = -5, because 5 2 = 25, (-5) 2 = 25

Square Root Function f(x) = x f(x) = x x y = x (x, y) 00(0,0) 11(1,1) 42(4,2) 93(9,3) 164(16,4) (18,4.2) x y

Evaluating a Square Root Function Given: f(x) = 12x – 20 Given: f(x) = 12x – 20 Find: f(3) Find: f(3) Solution: Solution: f(3) = 12(3) – 20 = 36 – 20 = 16 = 4 f(3) = 12(3) – 20 = 36 – 20 = 16 = 4

Domain of a Square Root Function Given: f(x) = 3x + 12 Given: f(x) = 3x + 12 Find the Domain of f(x): Find the Domain of f(x): Solution: Solution: 3x + 12 ≥ 0 3x ≥ -12 X ≥ -4 [-4, ∞) 3x + 12 ≥ 0 3x ≥ -12 X ≥ -4 [-4, ∞)

Application By 2005, an “hour-long” show on prime time TV was 45.4 min on the average, and the rest was commercials, plugs, etc. But this amount of “clutter“ was leveling off in recent years. The amount of non-program “clutter”, in minutes, was given by: M(x) = 0.7 x where x is the number of years after By 2005, an “hour-long” show on prime time TV was 45.4 min on the average, and the rest was commercials, plugs, etc. But this amount of “clutter“ was leveling off in recent years. The amount of non-program “clutter”, in minutes, was given by: M(x) = 0.7 x where x is the number of years after What was the number of minutes of “clutter” in an hour program in 2002? What was the number of minutes of “clutter” in an hour program in 2002?

Solution Solution: Solution: M(x) = 0.7 x M(x) = 0.7 x x = 2002 – 1996 = 6 M(6) = ~ 0.7(2.45) ~ 14.2 (min) x = 2002 – 1996 = 6 M(6) = ~ 0.7(2.45) ~ 14.2 (min) In 2009? In 2009? x = 2009 – 1996 = 13 x = 2009 – 1996 = 13 M(13) = ~ 15 (min) M(13) = ~ 15 (min)

Cube Root and Cube Root Function a = b, a = b, means b 3 = a means b 3 = a 8 = 2, 8 = 2, because 2 3 = 8 because 2 3 = = = -4 Because (-4) 3 = -64 Because (-4) 3 =

Cube Root Function f(x) = x f(x) = x 3 x y = x (x, y) -27-3(-27,-3) -8-2(-8,-2)(-1,-1) 00(0,0) 11(1,1) 82(8,2) 273(27,3) 303.1(30,3.1) 3

Simplifying Radical Expressions -64x 3 = (-4x) 3 = -4x -64x 3 = (-4x) 3 = -4x 81 = (3) 4 = 3 81 = (3) 4 = = x has no solution in R, -81 = x has no solution in R, since there is no x such that x 4 = -81 since there is no x such that x 4 = -81 In general In general -a has an nth root when n is odd -a has an nth root when n is odd -a has no nth root when n is even -a has no nth root when n is even n n

7.2 Rational Exponents What is the meaning of 7 1/3 ? What is the meaning of 7 1/3 ? x = 7 1/3 means x = 7 1/3 means X 3 = (7 1/3 ) 3 = 7 X 3 = (7 1/3 ) 3 = 7 Generally, a 1/n is number such that Generally, a 1/n is number such that (a 1/n ) n = a (a 1/n ) n = a

Your Turn Simplify Simplify /2 2. (-125) 1/3 3. (6x2y) 1/3 4. (-8) 1/3  Solutions x2y

Solve / /3 = (1000 1/3 ) 2 = 10 2 = 100 = (1000 1/3 ) 2 = 10 2 = /2 16 3/2 (16 1/2 ) 3 = 4 3 = 64 (16 1/2 ) 3 = 4 3 = / /5 -(32 1/5 ) 3 = -(2) 3 = -8 -(32 1/5 ) 3 = -(2) 3 = -8

Your Turn What is the difference What is the difference between -32 3/5 and (-32) 3/5 between -32 3/5 and (-32) 3/5 between -16 3/4 and (-16) 3/4 between -16 3/4 and (-16) 3/4

Simplify 6 1/7 · 6 4/7 6 1/7 · 6 4/7 = 6 (1/4 + 4/7) = 6 5/7 = 6 (1/4 + 4/7) = 6 5/7 32x 1/ x 3/4 32x 1/ x 3/4 = 2x (1/2 – 3/4) = 2x -1/4 = 2x (1/2 – 3/4) = 2x -1/4 (8.3 3/4 ) 2/3 (8.3 3/4 ) 2/3 = 8.3 (3/4 ∙ 2/3) = 8.3 1/2 = 8.3 (3/4 ∙ 2/3) = 8.3 1/2

Simplify 49 -1/ /2 = (7 2 ) -1/2 = 7 -1 = 1/7 = (7 2 ) -1/2 = 7 -1 = 1/7 (8/27) -1/3 (8/27) -1/3 = 1/(8/27) 1/3 = (27/8) 1/3 = 27 1/3 /8 1/3 = 3/2 = 1/(8/27) 1/3 = (27/8) 1/3 = 27 1/3 /8 1/3 = 3/2 (-64) -2/3 (-64) -2/3 = 1/(-64) 2/3 = 1/((-64) 1/3 ) 2 = 1/(-4) 2 = 1/16 = 1/(-64) 2/3 = 1/((-64) 1/3 ) 2 = 1/(-4) 2 = 1/16 (5 2/3 ) 3 (5 2/3 ) 3 = 5 2/3 ∙ 3 = 5 2 = 25 = 5 2/3 ∙ 3 = 5 2 = 25 (2x 1/2 ) 5 (2x 1/2 ) x 1/2 · 5 = 32x 5/2 2 5 x 1/2 · 5 = 32x 5/2

7.3 Multiplying & Simplifying Radical Expressions Product Rule Product Rule a · b = ab or a · b = ab or a 1/n · b 1/n = (ab) 1/n a 1/n · b 1/n = (ab) 1/n Note: Factors have same order of root. Note: Factors have same order of root. E.g, E.g, 25 4 = 25 · 4 = 100 = = 25 · 4 = 100 = = 400 · 5 = 400 · 5 = = 400 · 5 = 400 · 5 = 20 5 nn n nn

Simplify Radicals by Factoring √(80) √(80) = √(8 · 2 · 5) = √(2 3 · 2 · 5) = √(2 4 · 5) = 4√(5) = √(8 · 2 · 5) = √(2 3 · 2 · 5) = √(2 4 · 5) = 4√(5) √(40) √(40) = √(8 · 5) = √(2 3 · 5) = 2√(5) = √(8 · 5) = √(2 3 · 5) = 2√(5) √(200x 4 y 2 ) √(200x 4 y 2 ) = √(5 · 40x 4 y 2 ) = √(5 · 5 · 8x 4 y 2 ) = √(5 2 · 2 2 · 2x 4 y 2 ) = 5 · 2 x 2 y√(2) = 10x 2 y√(2) = √(5 · 40x 4 y 2 ) = √(5 · 5 · 8x 4 y 2 ) = √(5 2 · 2 2 · 2x 4 y 2 ) = 5 · 2 x 2 y√(2) = 10x 2 y√(2) √(80) √(80) = √(8 · 2 · 5) = √(2 3 · 2 · 5) = √(2 4 · 5) = 4√(5) = √(8 · 2 · 5) = √(2 3 · 2 · 5) = √(2 4 · 5) = 4√(5) √(40) √(40) = √(8 · 5) = √(2 3 · 5) = 2√(5) = √(8 · 5) = √(2 3 · 5) = 2√(5) √(200x 4 y 2 ) √(200x 4 y 2 ) = √(5 · 40x 4 y 2 ) = √(5 · 5 · 8x 4 y 2 ) = √(5 2 · 2 2 · 2x 4 y 2 ) = 5 · 2 x 2 y√(2) = 10x 2 y√(2) = √(5 · 40x 4 y 2 ) = √(5 · 5 · 8x 4 y 2 ) = √(5 2 · 2 2 · 2x 4 y 2 ) = 5 · 2 x 2 y√(2) = 10x 2 y√(2)

Simplify Radicals by Factoring √(64x 3 y 7 z 29 ) √(64x 3 y 7 z 29 ) = √(32 · 2x 3 y 5 y 2 z 25 z 4 ) = √(2 5 y 5 z 25 · 2x 3 y 2 z 4 ) = 2yz 5 √(2x 3 y 2 z 4 ) = √(32 · 2x 3 y 5 y 2 z 25 z 4 ) = √(2 5 y 5 z 25 · 2x 3 y 2 z 4 ) = 2yz 5 √(2x 3 y 2 z 4 )

Multiplying & Simplifying √(15) · √(3) √(15) · √(3) = √(45) = √(9 · 5) = 3√(5) = √(45) = √(9 · 5) = 3√(5) √(8x 3 y 2 ) · √(8x 5 y 3 ) √(8x 3 y 2 ) · √(8x 5 y 3 ) = √(64x 8 y 5 ) = √(16 · 4x 8 y 4 y) = 2x 2 y√(4y) = √(64x 8 y 5 ) = √(16 · 4x 8 y 4 y) = 2x 2 y√(4y)

Application Paleontologists use the function W(x) = 4 √(2x) to estimate the walking speed of a dinosaur, W(x), in feet per second, where x is the length, in feet, of the dinosaur ’ s leg. What is the walking speed of a dinosaur whose leg length is 6 feet? Paleontologists use the function W(x) = 4 √(2x) to estimate the walking speed of a dinosaur, W(x), in feet per second, where x is the length, in feet, of the dinosaur ’ s leg. What is the walking speed of a dinosaur whose leg length is 6 feet?

W(x) = 4 √(2x) W(x) = 4 √(2x) W(6) = 4√(2 · 6) = 4√(12) = 4√(4 · 3) = 8√(3) ~ 8√(1.7) ~ 14 (ft/sec) (humans: 4.4 ft/sec walking 22 ft/sec running) W(6) = 4√(2 · 6) = 4√(12) = 4√(4 · 3) = 8√(3) ~ 8√(1.7) ~ 14 (ft/sec) (humans: 4.4 ft/sec walking 22 ft/sec running)

Your Turn Simplify the radicals Simplify the radicals √(2x/3) · √(3/2) √(2x/3) · √(3/2) = √((2x/3)(3/2)) = √x = √((2x/3)(3/2)) = √x √(x/3) · √(7/y) √(x/3) · √(7/y) = √((x/3)(7/y)) = √(7x/3y) = √((x/3)(7/y)) = √(7x/3y) √(81x 8 y 6 ) √(81x 8 y 6 ) = √(27 · 3x 6 x 2 y 6 )= 3x 2 y 2 √(3x 2 ) = √(27 · 3x 6 x 2 y 6 )= 3x 2 y 2 √(3x 2 ) √((x+y) 4 ) √((x+y) 4 ) =√((x+y) 3 (x+y))= (x+y)√(x+y) =√((x+y) 3 (x+y))= (x+y)√(x+y)

7.4 Adding, Subtracting, & Dividing Adding (radicals with same indices & radicands) 8√(13) + 2√(13) 8√(13) + 2√(13) = √(13) · (8 + 2) = 10√(13) = √(13) · (8 + 2) = 10√(13) 7√(7) – 6x√(7) + 12√(7) 7√(7) – 6x√(7) + 12√(7) = √(7) · (7 – 6x + 12) = (19 – 6x)√(7) = √(7) · (7 – 6x + 12) = (19 – 6x)√(7) 7√(3x) - 2√(3x) + 2x 2 √(3x) 7√(3x) - 2√(3x) + 2x 2 √(3x) = √(3x) · (7 – 2 + 2x 2 ) = (5 + 2x 2 ) √(3x) = √(3x) · (7 – 2 + 2x 2 ) = (5 + 2x 2 ) √(3x)

Adding 7√(18) + 5√(8) 7√(18) + 5√(8) = 7√(9 · 2) + 5√(4 · 2) = 7 · 3 √(2) + 5 · 2√(2) = 21√(2) + 10√(2) = 31√(2) = 7√(9 · 2) + 5√(4 · 2) = 7 · 3 √(2) + 5 · 2√(2) = 21√(2) + 10√(2) = 31√(2) √(27x) - 8√(12x) √(27x) - 8√(12x) = √(9 · 3x) - 8√(4 · 3x) = 3√(3x) – 8 · 2√(3x) = √(3x) · (3 – 16) = -13√(3x) = √(9 · 3x) - 8√(4 · 3x) = 3√(3x) – 8 · 2√(3x) = √(3x) · (3 – 16) = -13√(3x) √(xy 2 ) + √(8x 4 y 5 ) √(xy 2 ) + √(8x 4 y 5 ) = √(xy 2 ) + √(8x 3 y 3 xy 2 ) = √(xy 2 ) + 2xy √(xy 2 ) = √(xy 2 ) (1 + 2xy) = (1 + 2xy) √(xy 2 ) = √(xy 2 ) + √(8x 3 y 3 xy 2 ) = √(xy 2 ) + 2xy √(xy 2 ) = √(xy 2 ) (1 + 2xy) = (1 + 2xy) √(xy 2 )

Dividing Radical Expressions Recall: (a/b) 1/n = (a) 1/n /(b) 1/n Recall: (a/b) 1/n = (a) 1/n /(b) 1/n (x 2 /25y 6 ) 1/2 (x 2 /25y 6 ) 1/2 =(x 2 ) 1/2 / (25y 6 ) 1/2 =x/5y 3 =(x 2 ) 1/2 / (25y 6 ) 1/2 =x/5y 3 ( 45xy ) 1/2 /( 2·5 1/2 ) ( 45xy ) 1/2 /( 2·5 1/2 ) = (1/2) · (45xy/5) 1/2 = (1/2) ·(9·5xy/5) 1/2 = (1/2) ·3(xy) 1/2 = (3/2) · (xy) 1/2 = (1/2) · (45xy/5) 1/2 = (1/2) ·(9·5xy/5) 1/2 = (1/2) ·3(xy) 1/2 = (3/2) · (xy) 1/2 (48x 7 y) 1/3 /(6xy -2 ) 1/3 (48x 7 y) 1/3 /(6xy -2 ) 1/3 = ((48x 7 y)/6xy -2 )) 1/3 = (8x 6 y 3 ) 1/3 = 2x 2 y = ((48x 7 y)/6xy -2 )) 1/3 = (8x 6 y 3 ) 1/3 = 2x 2 y

7.5 Rationalizing Denominators Given: 1 √(3) Rationalize the denominator—get rid of the radical in the denominator. Given: 1 √(3) Rationalize the denominator—get rid of the radical in the denominator. 1 √(3) √(3) = √(3) √(3) 3 1 √(3) √(3) = √(3) √(3) 3

Denominator Containing 2 Terms Given: 8 3√(2) + 4 Given: 8 3√(2) + 4 Rationalize denominator Rationalize denominator Recall: (A + B)(A – B) = A 2 – B 2 Recall: (A + B)(A – B) = A 2 – B 2 8 3√(2) – 4 8(3√(2) – 4) = 3√(2) + 4 3√(2) – 4 (3√(2) ) 2 – (4) 2 24 √(2) (3 √(2) – 4) 12 √(2) - 16 = = 18 – √(2) – 4 8(3√(2) – 4) = 3√(2) + 4 3√(2) – 4 (3√(2) ) 2 – (4) 2 24 √(2) (3 √(2) – 4) 12 √(2) - 16 = = 18 – 16 2

Your Turn Rationalize the denominator Rationalize the denominator 2 + √(5) √(6) - √(3) 2 + √(5) √(6) - √(3) 2+√(5) √(6)+√(3) 2√(6)+2√(3)+√(5)√(6)+√(5)√(3) = √(6) - √(3) √(6)+√(3) 6 – 3 2√(6) + 2√(3) + √(30) +√(15) = 3 2+√(5) √(6)+√(3) 2√(6)+2√(3)+√(5)√(6)+√(5)√(3) = √(6) - √(3) √(6)+√(3) 6 – 3 2√(6) + 2√(3) + √(30) +√(15) = 3

7.6 Radical Equations Application Application A basketball player’s hang time is the time in the air while shooting a basket. It is related to the vertical height of the jump by the following formula: t = √(d) / 2 A Harlem Globetrotter slam-dunked while he was in the air for 1.16 seconds. How high did he jump? A basketball player’s hang time is the time in the air while shooting a basket. It is related to the vertical height of the jump by the following formula: t = √(d) / 2 A Harlem Globetrotter slam-dunked while he was in the air for 1.16 seconds. How high did he jump?

Solving Radical Equations √(x) = 10 √(x) = 10 (√(x)) 2 = 10 2 x = 100 (√(x)) 2 = 10 2 x = 100 √(2x + 3) = 5 √(2x + 3) = 5 (√(2x + 3) ) 2 = 5 2 (2x + 3) = 25 2x = 22 x = 11 (√(2x + 3) ) 2 = 5 2 (2x + 3) = 25 2x = 22 x = 11 Check √(2x + 3) = 5 √(2(11) + 3) = 5 ? √(22 + 3) √(25) = 5 ? 5 = 5 yes √(2x + 3) = 5 √(2(11) + 3) = 5 ? √(22 + 3) = 5 ? √(25) = 5 ? 5 = 5 yes

Solve √(x - 3) + 6 = 5 √(x - 3) + 6 = 5 √(x - 3) = -1 (√(x - 3))2 = (-1)2 (x – 3) = 1 x = 4 √(x - 3) = -1 (√(x - 3))2 = (-1)2 (x – 3) = 1 x = 4 Check: √(x - 3) + 6 = 5 √(4 - 3) + 6 = 5 ? √(1) + 6 = 5 ? = 5 ? False √(x - 3) + 6 = 5 √(4 - 3) + 6 = 5 ? √(1) + 6 = 5 ? = 5 ? False Thus, there is no solution to this equation. Thus, there is no solution to this equation.

Your Turn Solve: √(x – 1) + 7 = 2 Solve: √(x – 1) + 7 = 2 √(x – 1) = -5 (√(x – 1)) 2 = (-5) 2 x – 1 = 25 x = 26 √(x – 1) = -5 (√(x – 1)) 2 = (-5) 2 x – 1 = 25 x = 26 Check: √(x – 1) + 7 = 2 √(26 – 1) + 7 = 2 ? √(25) + 7 = 2 ? = 2 ? False Thus, there is no solution to this equation.

Your Turn Solve: x + √(26 – 11x) = 4 Solve: x + √(26 – 11x) = 4 √(26 – 11x) = 4 – x (√(26 – 11x)) 2 = (4 – x) 2 26 – 11x = 16 – 8x + x 2 0 = x 2 + 3x – 10 x 2 + 3x – 10 = 0 (x – 2)(x + 5) = 0 x – 2 = 0 x = 2 x + 5 = 0 x = -5 √(26 – 11x) = 4 – x (√(26 – 11x)) 2 = (4 – x) 2 26 – 11x = 16 – 8x + x 2 0 = x 2 + 3x – 10 x 2 + 3x – 10 = 0 (x – 2)(x + 5) = 0 x – 2 = 0 x = 2 x + 5 = 0 x = -5 Check -5: √(26 – 11x) = 4 – x √(26 – 11(-5)) = 4 – (-5) ? √( ) = ? √(81) = 9 ? 9 = 9 True Check 2: √(26 – 11x) = 4 – x √(26 – 11(2)) = 4 – 2 ? √(4) = 2 ? 2 = 2 True Solution: {-5, 2}

Hang Time in Basketball A basketball player’s hang time is the time spent in the air when shooting a basket. It is a function of vertical height of jump. √(d) t = where t is hang time in sec and 2 d is vertical distance in feet. A basketball player’s hang time is the time spent in the air when shooting a basket. It is a function of vertical height of jump. √(d) t = where t is hang time in sec and 2 d is vertical distance in feet. If Michael Wilson of Harlem Globetrotters had a hang time of 1.16 sec, what was his vertical jump? If Michael Wilson of Harlem Globetrotters had a hang time of 1.16 sec, what was his vertical jump?

Hang Time √(d) t = t = √(d) 2(1.16) = √(d) 2.32 = √(d) (2.32) 2 = (√(d)) = d √(d) t = t = √(d) 2(1.16) = √(d) 2.32 = √(d) (2.32) 2 = (√(d)) = d

7.7 Complex Numbers What kind of number is x = √(-25)? What kind of number is x = √(-25)? x 2 = -25? x 2 = -25? Imaginary Unit i Imaginary Unit i i = √(-1), i 2 = -1 i = √(-1), i 2 = -1 Example Example √(-25) = √((25)(-1)) = √(25)√(-1) = 5i √(-25) = √((25)(-1)) = √(25)√(-1) = 5i √(-80) = √((80)(-1)) = √((16 · 5)(-1)) = 4√(5)i = 4i √(5) √(-80) = √((80)(-1)) = √((16 · 5)(-1)) = 4√(5)i = 4i √(5)

Your Turn Express the following with i. Express the following with i. 1.√(-49) 2.√(-21) 3.√(-125) 4.-√(-300)

Complex Numbers Comlex number has a Real part and an Imaginary part of the form: a + bi Comlex number has a Real part and an Imaginary part of the form: a + bi Example Example i i 3.5 – 2i

Adding and Subtracting Complex Numbers (5 – 11i) + (7 + 4i) = 5 – 11i i = 12 – 7i (5 – 11i) + (7 + 4i) = 5 – 11i i = 12 – 7i (2 + 6i) – (12 – 4i) = 2 + 6i – i = i (2 + 6i) – (12 – 4i) = 2 + 6i – i = i

Multiplying Complex Numbers 4i(3 – 5i) = 12i – 20i 2 = 12i – 20(-1) = i 4i(3 – 5i) = 12i – 20i 2 = 12i – 20(-1) = i (5 + 4i)(6 – 7i) = 5·6 – 5 ·7i + 4i· 6 – 4 ·7i 2 = 30 – 35i + 24i – 28(-1) = 30 – 11i + 28 = 58 – 11i (5 + 4i)(6 – 7i) = 5·6 – 5 ·7i + 4i· 6 – 4 ·7i 2 = 30 – 35i + 24i – 28(-1) = 30 – 11i + 28 = 58 – 11i

Multiplying 1.√(-3) √(-5) = i√(3) · i√(5) = i 2 √(15) = -√(15) 2.√(-5) √(-10) = i√(5) · i√(10) = i 2 √(50) = -√(50) = -√(25 · 2) = -5√(2)

Conjugates and Division Given: a + bi Conjugate of a + bi: a – bi Conjugate of a – bi: a + bi Given: a + bi Conjugate of a + bi: a – bi Conjugate of a – bi: a + bi Why conjugates? (a + bi)(a – bi) = (a) 2 – (bi) 2 = a 2 – b 2 i 2 = a 2 + b 2 Why conjugates? (a + bi)(a – bi) = (a) 2 – (bi) 2 = a 2 – b 2 i 2 = a 2 + b 2 (3 + 2i)(3 – 2i) = 9 – (2i) 2 = 9 – 4(-1) = 13 (3 + 2i)(3 – 2i) = 9 – (2i) 2 = 9 – 4(-1) = 13 Multiplying a complex number by its conjugate results in a real number. Multiplying a complex number by its conjugate results in a real number.

Dividing Complex Numbers Express 7 + 4i as a + bi 2 – 5i Express 7 + 4i as a + bi 2 – 5i 7 + 4i (7 + 4i) (2 + 5i) i + 8i = · = – 5 i (2 – 5i) (2 + 5i) – 43i = i (7 + 4i) (2 + 5i) i + 8i = · = – 5 i (2 – 5i) (2 + 5i) – 43i =

Your Turn 6 + 2i – 3i 6 + 2i – 3i 6 + 2i (4 + 3i) i + 8i + 6i 2 = · = (4 – 3i) (4 + 3i) ( i) = i (4 + 3i) i + 8i + 6i 2 = · = (4 – 3i) (4 + 3i) ( i) =

Your Turn 5i – i 5i – i (5i – 4) -3i -15i i = · = i -3i -9i i 3(5 + 4i) 5 + 4i = = = (5i – 4) -3i -15i i = · = i -3i -9i i 3(5 + 4i) 5 + 4i = = =

Powers of i i 2 = -1 i 3 = (-1)i = -i i 4 = (-1) 2 = 1 i 5 = (i 4 )i = i i 6 = (-1) 3 = -1 i 7 = (i 6 )i = -i i 8 = (-1) 4 = 1 i 9 = (i 8 )i = i i 10 = (-1) 5 = -1 i 2 = -1 i 3 = (-1)i = -i i 4 = (-1) 2 = 1 i 5 = (i 4 )i = i i 6 = (-1) 3 = -1 i 7 = (i 6 )i = -i i 8 = (-1) 4 = 1 i 9 = (i 8 )i = i i 10 = (-1) 5 = -1

Your Turn Simplify Simplify i 17 i 17 i 17 = i 16 i = (i 2 ) 8 i = i i 17 = i 16 i = (i 2 ) 8 i = i i 50 i 50 i 50 = (i 2 ) 25 = (-1) 25 = -1 i 50 = (i 2 ) 25 = (-1) 25 = -1 i 35 i 35 i 35 = (i 34 )i = (i 2 ) 17 i = (-1) 17 i = -i i 35 = (i 34 )i = (i 2 ) 17 i = (-1) 17 i = -i

Application Electrical engineers use the Ohm’s law to relate the current (I, in amperes), voltage (E, in volts), and resistence (R, in ohms) in a circuit: E = IR Electrical engineers use the Ohm’s law to relate the current (I, in amperes), voltage (E, in volts), and resistence (R, in ohms) in a circuit: E = IR Given: I = (4 – 5i) and R = (3 + 7i), what is E? Given: I = (4 – 5i) and R = (3 + 7i), what is E? E = (4 – 5i)(3 + 7i) = i - 15i - 35i 2 = i (volts) E = (4 – 5i)(3 + 7i) = i - 15i - 35i 2 = i (volts)