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Need help with complex integral

  • Thread starter Samuelb88
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  • #1
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Homework Statement


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].


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


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.
 

Answers and Replies

  • #2
CompuChip
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Are you sure that [itex]z_0 = \pi / 2[/itex] is inside your curve [itex]z(t) = e^{it}[/itex]?
 
  • #3
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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.
 
  • #4
CompuChip
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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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