Let Find с S = 2=-t P(2) q(z) be the 2²-1 Sinz poles inside the circle By residue theorem 250i/Res 22-1 Sinz = dz __2²_1 circle Sinz + Res 2²-1 Z=0 Sinz |z 2Ji = 2,5 i (1- JT ³² -1 + 1 - JT ²) - (2 ST - 4st ³) (₂ 3 Анд ore + residue → _p(2) 9'(2) at z =- J, 0, JT Res 2=J ((-³) -1 + = +*+ ² =1) Z²-1 Sinz cosz
Let Find с S = 2=-t P(2) q(z) be the 2²-1 Sinz poles inside the circle By residue theorem 250i/Res 22-1 Sinz = dz __2²_1 circle Sinz + Res 2²-1 Z=0 Sinz |z 2Ji = 2,5 i (1- JT ³² -1 + 1 - JT ²) - (2 ST - 4st ³) (₂ 3 Анд ore + residue → _p(2) 9'(2) at z =- J, 0, JT Res 2=J ((-³) -1 + = +*+ ² =1) Z²-1 Sinz cosz
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 38E
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![Let
Find
с
S
P(2)
q(z)
be the
2²-1
Sinz
=
poles inside the circle
By residue theorem
250i/Resz
Z=-t
2-1
Sinz
__2²_1
dz
Sinz
circle
+ Res 2²-1
Z=0
Sinz
|z
ore
2Ji
((-3) -1 + =
+
= 2,5T (1-JT ³² - 1 + 1 - JT ²)
- (2 st - 4st ³) (₂
3
Анд
+
residue → _p(2)
9'(2)
at 2 = -Jt₂0, JT
Res
2=J
+ 3+² -1)
2
Z²--1
Sinz
cosz](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa62f7b35-7db6-46d8-92c3-a45ad2747ea7%2F33c33a29-80f0-44d2-bd9f-e33d8c1ba179%2Fvo72t5b_processed.png&w=3840&q=75)
Transcribed Image Text:Let
Find
с
S
P(2)
q(z)
be the
2²-1
Sinz
=
poles inside the circle
By residue theorem
250i/Resz
Z=-t
2-1
Sinz
__2²_1
dz
Sinz
circle
+ Res 2²-1
Z=0
Sinz
|z
ore
2Ji
((-3) -1 + =
+
= 2,5T (1-JT ³² - 1 + 1 - JT ²)
- (2 st - 4st ³) (₂
3
Анд
+
residue → _p(2)
9'(2)
at 2 = -Jt₂0, JT
Res
2=J
+ 3+² -1)
2
Z²--1
Sinz
cosz
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