MHB Contour integral (please check my solution)

aruwin
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Hello. Can someone check my solution for this question? I am not sure what to do about the "from e^-pi*1/2 to e^pi*i/2" part. I ignored that part.
 

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That looks correct. You could simplify the answer by noticing that $\sin\pi = 0.$
 
Opalg said:
That looks correct. You could simplify the answer by noticing that $\sin\pi = 0.$

Thanks. But what does "from e^-pi*1/2 to e^pi*i/2" mean?
 
aruwin said:
Thanks. But what does "from e^-pi*1/2 to e^pi*i/2" mean?
It means that $C$ is the contour starting at the point $e^{-i\pi/2} = -i$ on the unit circle, and ending at the point $e^{i\pi/2} = i.$
 
Opalg said:
It means that $C$ is the contour starting at the point $e^{-i\pi/2} = -i$ on the unit circle, and ending at the point $e^{i\pi/2} = i.$

Does that affect the calculation here? Is that why we integrate from -pi/2 to pi/2?
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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