Cauchy's Integral Formula problem

[paddo]

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

Cheers

 

[jostpuur]

Integration domain would be relevant.

 

[paddo]

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

 

[jostpuur]

The knowledge that the contour goes once around z=1 should be enough. The comment on point z=i looks like misdirection.

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?

 

[paddo]

Yeah. I know the formula.

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

 

[jostpuur]

There exists coefficients $a_{-2}, a_{-1}, a_0, a_1, \ldots$ so that

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

For integration, you need to know the $a_{-1}$.