# Need help with complex integral

## 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.

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

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.

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