Week 4 Worksheet - Solutions
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Contours
We calculate so that . Then
The following parameterization works.
A straight line from to .
The reverse of should be .
Contour Integration
- We take on so that and . Thus the integral is
- Here on so and . Also .
- Use on . Here and and .
- and
- and
- Take on . Then .
and the integral is
Antiderivatives
- We can get an antiderivative from integration by parts
and the fundamental theorem of contour integration lets us evaluate the contour integral easily.
- Integration by parts again helps with the antiderivative
and
- where on and on . Then
- where on and on . Then
- There is no possibility that has an antiderivative. If it did, then the contour integral of over the closed contour would be zero and our answers to a) and b) would have to be equal.
- on
- has an antiderivative on . Therefore
- In this case . The function still has an antiderivative, so
- The answer to c) is the negative of the answer to b)
- The function has an antiderivative by the product rule. Therefore
and subtracting the contour integral of gives the result.
The Principal Logarithm
- can take any real value and any imaginary value between and .
Therefore the range is .
- In order to calculate we need the absolute value and the principal argument of . Write so that .
- Take . Then and because we must use the principal argument, but this is not twice .
- Let and be the principal arguments of and respectively. We know from polar multiplication that is an argument of . It may not be the principal argument, but there is such that .
Thus
-
and , so .
and so .
- Take . Then but and .