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1 Given the Jordan curve y that is the counterclockwise oriented boundary of the set T= {z C |+1| = 1/2}, the complex line integral 1 √√√ ²(= + 1) (2 + 1 − 1) dz takes value 8πί (A) 3 128πί 15-8i -8πί 8π(1+81) 15-8i z² (z+1) (2+1 - 1/4) 2=0→> double pole → 10+11 > 1/2 → → outside z = -1 → single pole →1-1+1|< 1/2 inside single pole →1-1+1 +11 < 1/2 → inside z = -1+4 + lim (2+1). Z→-1 + 1 = 4i-8π 1(-114) + + lim 2-1+1/4 (211-114) z² (z+1) (2+1 1/4) (-1+1/4)² (-1+1/4++) (1/15 - 11/21) (1/4) 321Ti (15-81) (1/4)

1 Given the Jordan curve y that is the counterclockwise oriented boundary of the set T= {z C |+1| = 1/2}, the complex line integral 1 √√√ ²(= + 1) (2 + 1 − 1) dz takes value 8πί (A) 3 128πί 15-8i -8πί 8π(1+81) 15-8i z² (z+1) (2+1 - 1/4) 2=0→> double pole → 10+11 > 1/2 → → outside z = -1 → single pole →1-1+1|< 1/2 inside single pole →1-1+1 +11 < 1/2 → inside z = -1+4 + lim (2+1). Z→-1 + 1 = 4i-8π 1(-114) + + lim 2-1+1/4 (211-114) z² (z+1) (2+1 1/4) (-1+1/4)² (-1+1/4++) (1/15 - 11/21) (1/4) 321Ti (15-81) (1/4)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 12RE
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i circled the correct answer and i did most of the question but i cant figure out how to add both residues to get the correct answer

could you please show me how to do it

1 Given the Jordan curve y that is the counterclockwise oriented boundary of the set T= {z C |+1| = 1/2}, the complex line integral 1 √√√ ²(= + 1) (2 + 1 − 1) dz takes value 8πί (A) 3 128πί 15-8i -8πί 8π(1+81) 15-8i z² (z+1) (2+1 - 1/4) 2=0→> double pole → 10+11 > 1/2 → → outside z = -1 → single pole →1-1+1|< 1/2 inside single pole →1-1+1 +11 < 1/2 → inside z = -1+4 + lim (2+1). Z→-1 + 1 = 4i-8π 1(-114) + + lim 2-1+1/4 (211-114) z² (z+1) (2+1 1/4) (-1+1/4)² (-1+1/4++) (1/15 - 11/21) (1/4) 321Ti (15-81) (1/4)
Transcribed Image Text:1 Given the Jordan curve y that is the counterclockwise oriented boundary of the set T= {z C |+1| = 1/2}, the complex line integral 1 √√√ ²(= + 1) (2 + 1 − 1) dz takes value 8πί (A) 3 128πί 15-8i -8πί 8π(1+81) 15-8i z² (z+1) (2+1 - 1/4) 2=0→> double pole → 10+11 > 1/2 → → outside z = -1 → single pole →1-1+1|< 1/2 inside single pole →1-1+1 +11 < 1/2 → inside z = -1+4 + lim (2+1). Z→-1 + 1 = 4i-8π 1(-114) + + lim 2-1+1/4 (211-114) z² (z+1) (2+1 1/4) (-1+1/4)² (-1+1/4++) (1/15 - 11/21) (1/4) 321Ti (15-81) (1/4)
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