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| + | For VII.4.1 I get the bottom side of the rectangle tends to pi and the sides tend to 0. But when I try and evaluate the top side by letting <math>z (t) = 2at - a + \sqrt b i</math>, my integrand becomes a mess: <math>2a/((4a^2 t^2-4a^2 t+a^2-b+1)+i(4a\sqrt b t-2a\sqrt b))</math>. I don't see how that simplifies to what the integrand in the problem gives. Just keep messing around with the algebra until something happens? Or did I do something wrong | ||
| + | Edit: Never mind, I figured it out by changing the parametrization so <math>-a<=t<=a</math>--[[User:Aata|Adam Ata]] | ||
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| + | Just work with the algebra and then take the real part and you should get it | ||
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| + | --[[User:Kfernan|Kevin Fernandes]] 20:11, 29 October 2009 (UTC) | ||
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| + | Try multiplying by the complex conjugate. | ||
| + | --[[User:Ysuo|Yu Suo]] 20:40, 29 October 2009 (UTC) | ||
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| + | I'm confused as to how to start VII.4.2, could someone explain where to go with it? --[[User:Whoskins|Weston Hoskins]] | ||
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| + | For the portion of the curve on the real axis (z=t, 0<=t<=R), you integrate <math>1/(t^2+a^2)</math> from 0 to infinity. The integral of <math>1/(t^2+a^2)</math> is <math>arctan(t/a)/a</math>. | ||
| + | Next show that the integral of 'circle portion' is 0. | ||
| + | For the last part of the curve must be equal to negative of what you got in the first part because the sum of all 3 must equal 0. | ||
| + | <math>z = t*exp(\pi i/4)</math> for the last part. | ||
| + | Simplify the integral by multiplying by its complex conjugate and take real and imaginary parts. --[[User:Aata|Adam Ata]] | ||
Latest revision as of 05:06, 30 October 2009
Homework 7
For VII.4.1 I get the bottom side of the rectangle tends to pi and the sides tend to 0. But when I try and evaluate the top side by letting $ z (t) = 2at - a + \sqrt b i $, my integrand becomes a mess: $ 2a/((4a^2 t^2-4a^2 t+a^2-b+1)+i(4a\sqrt b t-2a\sqrt b)) $. I don't see how that simplifies to what the integrand in the problem gives. Just keep messing around with the algebra until something happens? Or did I do something wrong Edit: Never mind, I figured it out by changing the parametrization so $ -a<=t<=a $--Adam Ata
Just work with the algebra and then take the real part and you should get it
--Kevin Fernandes 20:11, 29 October 2009 (UTC)
Try multiplying by the complex conjugate.
--Yu Suo 20:40, 29 October 2009 (UTC)
I'm confused as to how to start VII.4.2, could someone explain where to go with it? --Weston Hoskins
For the portion of the curve on the real axis (z=t, 0<=t<=R), you integrate $ 1/(t^2+a^2) $ from 0 to infinity. The integral of $ 1/(t^2+a^2) $ is $ arctan(t/a)/a $. Next show that the integral of 'circle portion' is 0. For the last part of the curve must be equal to negative of what you got in the first part because the sum of all 3 must equal 0. $ z = t*exp(\pi i/4) $ for the last part. Simplify the integral by multiplying by its complex conjugate and take real and imaginary parts. --Adam Ata
