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
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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Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6389447-1237-4af0-b5c6-eb1260425b55%2F20f0b64d-66f5-4f4f-8f6f-529cbc96b077%2Fdi1t7cd_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6389447-1237-4af0-b5c6-eb1260425b55%2F20f0b64d-66f5-4f4f-8f6f-529cbc96b077%2Fcaqvvef_processed.jpeg&w=3840&q=75)
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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