Recent content by opticaltempest

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    Laurent Series for f(z)=(1+2z)/(z^2+z^3)

    Would anyone be willing to check and comment on my work for finding the Laurent series of f(z)=\frac{1+2z}{z^2+z^3} ? Page 1 - http://img23.imageshack.us/img23/7172/i0001.jpg" Page 2 - http://img5.imageshack.us/img5/2140/i0002.jpg" Page 3 -...
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    How can we obtain and manipulate Laurent series for different annuli?

    Well maybe since 1/(1-z) converges for |z|>1, with z=1/z, we get |1/z|<1 which is equivalent to |z|>1. Is multiplying the fraction by 1/z a standard procedure for finding the series that converges for |z|>1 ?
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    How can we obtain and manipulate Laurent series for different annuli?

    I uploaded a scanned page from Schaum's Outline of Complex Variables. I have some questions on how they found the Laurent series in Example 27. http://img19.imageshack.us/img19/7172/i0001.jpg If the image doesn't load, go to - http://img19.imageshack.us/img19/7172/i0001.jpg" In...
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    Is Cauchy's integral formula applicable to this type of integral?

    Ok, so it appears that I need to use the residue theorem in order to evaluate this integral. I was hoping I could just use the integral formula. I haven't got to study the residue theorem yet in my text. Thanks
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    Is Cauchy's integral formula applicable to this type of integral?

    Homework Statement I am trying to determine if Cauchy's integral formula will work on the following integral, where the contour C is the unit circle traversed in the counterclockwise direction. \oint_{C}^{}{\frac{z^2+1}{e^{iz}-1}} Homework Equations See Cauchy's Integral Formula -...
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    Question integrating |f(z)||dz| over a contour C

    Ok, I see why. Thank you very much!
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    Question integrating |f(z)||dz| over a contour C

    OK, I should have had |\gamma^{'}(t)|, instead of \gamma^{'}(t). So the original integral becomes \int_{0}^{1}{\sqrt{1+4t^2}}dt or would I have \int_{0}^{1}{\sqrt{1+4t^2} \cdot \sqrt{1+4t^2}dt} ?
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    Question integrating |f(z)||dz| over a contour C

    Homework Statement I want to compute \int_{C}^{}{|f(z)||dz|} along the contour C given by the curve y=x^2 using endpoints (0,0) and (1,1). I am to use f(z)=e^{i\cdot \texrm{arg}(z)} Homework Equations The Attempt at a Solution I know that for all complex numbers z, |e^{i\cdot...
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    Prove that if C and D are closed sets, then C U D is a closed set.

    Well, I think this is true. Intuitively, I want to say that if there is a sequence {xn} with some points in both C and D that accumulate at x0, and we remove those points only in D, then we would still be able to find subsequences in C that converge to x0, and vice versa. Regarding me not...
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    Prove that if C and D are closed sets, then C U D is a closed set.

    Does this look better? Proof: Suppose x_0 is an accumulation point of C \bigcup D. Then either x_0 is an accumulation point of C or x_0 is an accumulation point of D or x_0 is an accumulation point of both C and D. Case I: Without loss of generality, suppose that x_0 is an accumulation...
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    Prove that if C and D are closed sets, then C U D is a closed set.

    Ok, I see what I did wrong. Let me try this again. Thanks
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    Prove that if C and D are closed sets, then C U D is a closed set.

    How does this proof look? Proof: Suppose x_0 is an accumulation point of C \bigcup D. Then either x_0 is an accumulation point of C or x_0 is an accumulation point of D or x_0 is an accumulation point of both C and D. Case I: Without loss of generality, suppose that x_0 is an accumulation...
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    Prove that if C and D are closed sets, then C U D is a closed set.

    I wish this was trivial for me. :) Well, I had the correct intuition on how to prove the statement. Let me see if I can write the proof correctly. Thanks.
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    Prove that if C and D are closed sets, then C U D is a closed set.

    Homework Statement I want to show that if C and D and closed sets, then C \bigcup D is a closed set. Homework Equations A set is called closed iff the set contains all of its accumulation points. The Attempt at a Solution In order for me to prove this statement, I will be able...
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