2+2 -dz; = z²(z-1-i) (a) || = 1, (b) | - 1 - i = 1 (5) f 1 fc = 23 ( 2 - 4) (a) | = 1, (b) |z – 2 = 1 -dz;

College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter1: Equations And Inequalities
Section1.CR: Chapter Review
Problem 67E: 67. Calculate the discriminant associated with the equation 6x2+5x+1=0.
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Evaluate contour  integrals 

2+2
-dz;
z²(z - 1 - i)
(a) || = 1,
(b) | 2 - 1 - i = 1
(5) f
1
fo
z³(z - 4)
(a) || = 1,
(b) | = 2 = 1
=
- dz;
Transcribed Image Text:2+2 -dz; z²(z - 1 - i) (a) || = 1, (b) | 2 - 1 - i = 1 (5) f 1 fo z³(z - 4) (a) || = 1, (b) | = 2 = 1 = - dz;
Expert Solution
Step 1: ''Introduction to the solution''

5) a) Let I subscript 1 equals integral subscript C fraction numerator z plus 2 over denominator z squared open parentheses z minus 1 minus i close parentheses end fraction d z comma space w h e r e space C colon space open vertical bar z close vertical bar equals 1

                 equals integral subscript C space fraction numerator g left parenthesis z right parenthesis over denominator z squared end fraction d z....... left parenthesis 1 right parenthesis space comma w h e r e space g left parenthesis z right parenthesis equals fraction numerator left parenthesis z plus 2 right parenthesis over denominator open parentheses z minus 1 minus i close parentheses end fraction space

Cleary, g is  analytic inside and on C.

Then, we can apply the Cauchy Integral theorem for the above integral.

Cauchy Integral's theorem: Let f left parenthesis z right parenthesis  be an analytic  function inside and on a  closed contour C and alpha element ofIntC, being the interior of C. Then, f to the power of left parenthesis n right parenthesis end exponent left parenthesis alpha right parenthesis equals fraction numerator n factorial over denominator 2 pi i end fraction integral subscript C fraction numerator f left parenthesis z right parenthesis over denominator open parentheses z minus alpha close parentheses to the power of n plus 1 end exponent end fraction........ left parenthesis 2 right parenthesis comma n equals 0 comma 1 comma 2 comma......

Then,integral subscript C space fraction numerator g left parenthesis z right parenthesis over denominator z squared end fraction d z equals fraction numerator 2 pi i over denominator 1 factorial end fraction. g apostrophe left parenthesis 0 right parenthesis equals space 2 pi i cross times g apostrophe left parenthesis 0 right parenthesis......... left parenthesis 3 right parenthesis

Now,g apostrophe left parenthesis z right parenthesis equals fraction numerator left parenthesis z minus 1 minus i right parenthesis left parenthesis 1 right parenthesis minus left parenthesis z plus 2 right parenthesis left parenthesis 1 right parenthesis over denominator open parentheses z minus 1 minus i close parentheses squared end fraction equals fraction numerator negative left parenthesis 3 plus i right parenthesis over denominator open parentheses z minus 1 minus i close parentheses squared end fraction

 rightwards double arrow g to the power of apostrophe open parentheses 0 close parentheses equals negative fraction numerator open parentheses 3 plus i close parentheses over denominator open parentheses 1 plus i close parentheses squared end fraction equals negative fraction numerator open parentheses 3 plus i close parentheses over denominator left parenthesis 1 minus 1 plus 2 i right parenthesis end fraction equals negative fraction numerator left parenthesis 3 plus i right parenthesis over denominator 2 i end fraction equals negative 1 half open parentheses 1 plus 3 over i close parentheses equals negative 1 half open parentheses 1 minus 3 i close parentheses equals fraction numerator left parenthesis 3 i minus 1 right parenthesis over denominator 2 end fraction

Now,using this value  in  (3), we get, integral subscript C space fraction numerator g left parenthesis z right parenthesis over denominator z squared end fraction d z equals fraction numerator left parenthesis 3 i minus 1 right parenthesis over denominator 2 end fraction cross times 2 straight pi straight i equals straight pi straight i left parenthesis 3 straight i minus 1 right parenthesis

(b) Let I subscript 2 equals integral subscript C space fraction numerator f left parenthesis z right parenthesis over denominator open parentheses z minus 1 minus i close parentheses end fraction d z comma space w h e r e space f left parenthesis z right parenthesis equals fraction numerator left parenthesis z plus 2 right parenthesis over denominator left parenthesis z minus 1 minus i right parenthesis end fraction comma spacebeing analytic inside and on C: open vertical bar z minus left parenthesis 1 plus i right parenthesis close vertical bar equals 1

   Then, by Cauchy's  Integral theorem, I subscript 2 equals 2 straight pi straight i cross times straight f left parenthesis 1 plus straight i right parenthesis equals 2 straight pi straight i cross times fraction numerator left parenthesis 1 plus straight i plus 2 right parenthesis over denominator open parentheses 1 plus straight i close parentheses squared end fraction equals straight pi left parenthesis 3 plus straight i right parenthesis

             


                      



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