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2+1= (2-2) (2-2) (2-22)(Z-Z3)(Z-24) (2-25) Des 2-1 -1 (2-20) 22 Z=Z0 = = Z-Zo Z + 1 Z 2 li 2² +2(2-30)2 22 Z-020 6Z 67 685 い 3 630 Res + Res + Res 5=23.5=-2 S=è 12 623 623 影 Rese )= 53³+8 Hi S= 5=2+) = 2-2, 2 3=-8=8e F 3. F(s) = = 12 $³ +8* Suggestion: After finding the three cube roots -2 and 1±√√3i of -8, it is helpful to notice that the property z + z = 2 Re z of complex numbers enables one to write ei√√31 e-i√√31 = 2 Re ei√√3t -1+i√3 + -1+i√√3-1-i√√3 -2t Ans. f(t) = e² + e' (√√3 sin √√√3t - cos √√3t). COS

2+1= (2-2) (2-2) (2-22)(Z-Z3)(Z-24) (2-25) Des 2-1 -1 (2-20) 22 Z=Z0 = = Z-Zo Z + 1 Z 2 li 2² +2(2-30)2 22 Z-020 6Z 67 685 い 3 630 Res + Res + Res 5=23.5=-2 S=è 12 623 623 影 Rese )= 53³+8 Hi S= 5=2+) = 2-2, 2 3=-8=8e F 3. F(s) = = 12 $³ +8* Suggestion: After finding the three cube roots -2 and 1±√√3i of -8, it is helpful to notice that the property z + z = 2 Re z of complex numbers enables one to write ei√√31 e-i√√31 = 2 Re ei√√3t -1+i√3 + -1+i√√3-1-i√√3 -2t Ans. f(t) = e² + e' (√√3 sin √√√3t - cos √√3t). COS

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use method in the second image to solve this pro

2+1= (2-2) (2-2) (2-22)(Z-Z3)(Z-24) (2-25) Des 2-1 -1 (2-20) 22 Z=Z0 = = Z-Zo Z + 1 Z 2 li 2² +2(2-30)2 22 Z-020 6Z 67 685 い 3 630 Res + Res + Res 5=23.5=-2 S=è 12 623 623 影 Rese )= 53³+8 Hi S= 5=2+) = 2-2, 2 3=-8=8e F
Transcribed Image Text:2+1= (2-2) (2-2) (2-22)(Z-Z3)(Z-24) (2-25) Des 2-1 -1 (2-20) 22 Z=Z0 = = Z-Zo Z + 1 Z 2 li 2² +2(2-30)2 22 Z-020 6Z 67 685 い 3 630 Res + Res + Res 5=23.5=-2 S=è 12 623 623 影 Rese )= 53³+8 Hi S= 5=2+) = 2-2, 2 3=-8=8e F
3. F(s) = = 12 $³ +8* Suggestion: After finding the three cube roots -2 and 1±√√3i of -8, it is helpful to notice that the property z + z = 2 Re z of complex numbers enables one to write ei√√31 e-i√√31 = 2 Re ei√√3t -1+i√3 + -1+i√√3-1-i√√3 -2t Ans. f(t) = e² + e' (√√3 sin √√√3t - cos √√3t). COS
Transcribed Image Text:3. F(s) = = 12 $³ +8* Suggestion: After finding the three cube roots -2 and 1±√√3i of -8, it is helpful to notice that the property z + z = 2 Re z of complex numbers enables one to write ei√√31 e-i√√31 = 2 Re ei√√3t -1+i√3 + -1+i√√3-1-i√√3 -2t Ans. f(t) = e² + e' (√√3 sin √√√3t - cos √√3t). COS
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