- #1

chill_factor

- 901

- 5

## Homework Statement

Evaluate the integral ∫sin(z)/(z^2-4) dz about the contour C such that it is a circle of radius 2 centered at z = 2.

## Homework Equations

All theorems of complex analysis except residue theorem.

## The Attempt at a Solution

There is a singularity at z = 2, so we avoid it by drawing a bridge going from the outer contour now renamed C1, to another circular contour C2 with opposite orientation going around z = 2. The bridges have opposite orientation but same direction, so they cancel out, and the distance between them can be made arbitrarily small.

Therefore the summation of these 2 integrals is C, within which the function is analytic.

Then we recognize that ∫ C1 + ∫ C2 = ∫C = 0.

However I don't know how to apply this directly to solve the problem.

I tried a different way. Lets take the analytic contour C.

Note that f(z) = sin(z)/(z^2-4) = sin(z)/{ (z-2)(z+2) } = sin(z)(z+2)^-1/(z-2)

∫f(z)dz is now in the form ∫f(z)/(z-a) dz = 2πi f(a). Here a = 2, f(z) is now sin(z)(z+2)^-1

Substitute 2 into the equation and we have the integral around C is equal to

2πi*sin(2)/4 = πi sin(2)/2.

However this is for the entire contour C, while what I'm looking for is C1, the outer contour.

Can I just say that since C2 can be made arbitrarily small, the integral around it must be zero, and the answer I got for C is the final answer? If not what should my next step be?