Need help with complex integral


by Samuelb88
Tags: complex, integral
Samuelb88
Samuelb88 is offline
#1
Nov3-11, 12:45 AM
P: 162
1. The problem statement, all variables and given/known data
Let C be a loop around [itex]\pi/2[/itex]. Find the value of [itex]\frac{1}{2\pi i} \int_C \frac{\sin(z)}{(z-\pi/2)^3} dz[/itex].


2. Relevant 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 [itex]f^{(n)}(z_0) = \frac{(n-1)!}{2 \pi i} \int_C \frac{f(w)}{(w-z_0)^n} dw[/itex].


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

[itex]\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}[/itex]

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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CompuChip
CompuChip is offline
#2
Nov3-11, 02:19 AM
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Are you sure that [itex]z_0 = \pi / 2[/itex] is inside your curve [itex]z(t) = e^{it}[/itex]?
Samuelb88
Samuelb88 is offline
#3
Nov3-11, 09:40 AM
P: 162
Opps! I would need to use [itex]z(t) = \pi/2 + e^{it}[/itex], 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
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#4
Nov3-11, 04:57 PM
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Need help with complex integral


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?


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