How Is the Integral of Sin(z)/(z-pi/2)^3 Around a Loop Calculated?

[Samuelb88]

Homework Statement

Let C be a loop around \pi/2. Find the value of \frac{1}{2\pi i} \int_C \frac{\sin(z)}{(z-\pi/2)^3} dz.

Homework Equations

Thm: If f is analytic in its simply connected domain D, and C is a simply closed positively oriented loop that lies in D, and if z lies in the inside of C, then f^{(n)}(z_0) = \frac{(n-1)!}{2 \pi i} \int_C \frac{f(w)}{(w-z_0)^n} dw.

The Attempt at a Solution

Let f(z) = \sin(z) which is analytic for every z \in \mathbb{C}. We can parametrize C by z(t) = e^{it} and so C is a simply closed positively oriented curve. So I can apply my theorem to find the value of this integral. Hence:

\frac{1}{2 \pi i} \int_C \frac{\sin(z)}{(z-\pi/2)^3} dz = \frac{1}{2!} \frac{d^2}{dx^2} \sin(z) \Big|_{z=\pi/2} = -\frac{1}{2} \sin(\pi/2) = -\frac{1}{2}

I checked my answer against Wolfram Alpha which says the integral is equal to 0! Am I applying the theorem incorrectly? I can't figure out what's wrong.

 

[CompuChip]

Are you sure that z_0 = \pi / 2 is inside your curve z(t) = e^{it}?

 

[Samuelb88]

Opps! I would need to use z(t) = \pi/2 + e^{it}, but shouldn't I still be able to apply my theorem to find that the integral is equal to -1/2? Wolfram is still giving me an output of 0.

 

[CompuChip]

Hmm, you are right, I thought that it would give you sin(pi/2 + (pi/2 + 0)).

OK, let me try again: what is the value of n?