Recent content by gbu

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    Complex Laurent Series for (z+2)/(z-1) on Annulus Region

    How did I know to write it like that? It makes sense (and thanks a TON for your explanation of the annulus stuff)... I guess I just don't see how you go from the z+2/z-1 to the proper representation for the right series "on demand" so to speak.
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    Complex Laurent Series for (z+2)/(z-1) on Annulus Region

    OK, yes. Thank you! I recognize that... but a power series around what value? What do I do with the C_0 and C_1 annuli? A power series around 0 makes it pretty easy (though the answer doesn't appear sensical), but the C_0 says 1 < |z| and the |z| doesn't meet that condition at 0. Although...
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    Complex Laurent Series for (z+2)/(z-1) on Annulus Region

    Homework Statement Find the Laurent series of \frac{(z+2)}{(z-1)} on C_1: 1 < |z| and C_2: 0 < |z| < 1 Homework Equations I have a formula for computing Laurent series, but it includes an integral that is impossible to solve. For everything that I've read, no one actually...
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    Complex Variables Algebra Solutions / Argument/Modulus

    It's \sqrt{(4x^2 y^2 + (x^2-y^2-1)^2)} Which I suppose gets me to an answer, but its certainly not a pretty one. Wolfram Alpha gives a very simple answer to the question (a^2 = sqrt(1+i))
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    Complex Variables Algebra Solutions / Argument/Modulus

    Homework Statement Solve for a, a \in \mathbb{C} \frac{2\ln(a^2 - 1)}{\pi i} = 1 Homework Equations N/A.The Attempt at a Solution Reorganizing the equation. 2\log(a^2 - 1) = \pi i
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    Complex Analysis - Transcendental Solutions Help

    Brilliant, thank you jackmell! That's exactly what I was missing. It's funny, the textbook clearly says that, but I just didn't really "see" it.
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    Complex Analysis - Transcendental Solutions Help

    Right. Which leaves me with e^{-2b} = 1 Which we can log both sides... -2b = \ln 1 = 0 Telling me b = 0. As for a, we have e^{-2b}(\cos(2a) + i\sin(2a)) = 1, but we know b is 0. So we have [itex ]\cos(2a) + i\sin(2a) = 1[/itex] [itex ]\cos(2a) + i\sin(2a) = 1[/itex] or [itex ]e^{2ia}...
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    Complex Analysis - Transcendental Solutions Help

    I can see intuitively the correct answer from the first line (thanks a lot for that!), it makes it obvious that n \pi is an answer... I'm still not sure how to solve a and b in that equation though.e^{-2b}(\cos(2a)+i\sin(2a)) = 1 \cos(2a)+i\sin(2a) = e^{2b} I don't know where to go from here...
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    Complex Analysis - Transcendental Solutions Help

    This isn't really homework help. I'm working through a complex analysis textbook myself, and am stumped on the complex transcendentals, but I figured this was the best place for it. I would greatly appreciate any guidance here, I'm getting very frustrated! Homework Statement The problem is to...
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