Author Topic: 2.3 question 3  (Read 386 times)

Yan Zhou

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2.3 question 3
« on: February 10, 2020, 07:22:49 PM »
$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z} $$
Since -2 lies in $|z+1| = 2$, Cauchy theorem gives that
$$\frac{1}{2\pi i} \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{z-(-2)} = \frac{-2}{2-(-2)} = -\frac{1}{2}$$
then $$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = -\pi i$$
However, I checked the answer of textbook and it says the answer is $2\pi i$, I am confused about where i did wrong.

Yan Zhou

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Re: 2.3 question 3
« Reply #1 on: February 10, 2020, 07:46:22 PM »
I see. The question on the home assignment is a little bit different from the textbook.

Yan Zhou

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Re: 2.3 question 3
« Reply #2 on: February 10, 2020, 08:11:40 PM »
I found some differences between textbook and home assignment, some of them are typos, and some do not affect the questions but the answers. Should we always follow the questions on the textbook?

Victor Ivrii

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Re: 2.3 question 3
« Reply #3 on: February 11, 2020, 07:37:38 AM »
Your answer is correct ($-\pi i$).

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