- #1

- 11

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How would you integrate sin(z)/(z-1)^2 using Cauchy's Integral Formula? 1 is in C.

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Cheers

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- Thread starter paddo
- Start date

- #1

- 11

- 0

How would you integrate sin(z)/(z-1)^2 using Cauchy's Integral Formula? 1 is in C.

Cheers

Cheers

- #2

- 2,112

- 18

Integration domain would be relevant.

- #3

- 11

- 0

All it says is that "C is any simple closed contour around both z = 1 and z = i"

- #4

- 2,112

- 18

I believe that actually you already know what you want there, assuming that you know the Cauchy's integral formula. It's just that the 1/(z-1)^2 is confusing?

- #5

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Yeah. I know the formula.

I did 1/(z-1)^2 but didn't come out as partial fractions.

I did 1/(z-1)^2 but didn't come out as partial fractions.

- #6

- 2,112

- 18

[tex]

\frac{\sin z}{(z-1)^2} = \frac{a_{-2}}{(z-1)^2} \;+\; \frac{a_{-1}}{z-1} \;+\; a_0 \;+\; a_1(z-1) \;+\; \cdots

[/tex]

For integration, you need to know the [itex]a_{-1}[/itex].

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