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 $--Aata
Just work with the algebra and then take the real part and you should get it
--Kfernan 20:11, 29 October 2009 (UTC)
Try multiplying by the complex conjugate.
--Ysuo 20:40, 29 October 2009 (UTC)
The only part of 4.1 I'm stuck on is showing the bottom segment tends to pi. I parameterized it as z(t)=t, and that yields that the integral is arctan(t) evaluated between -a and a. Do I have to express arctan in terms of log to show that it tends to pi as 'a' tends to infinity?
--Dgoodin 22:02, 29 October 2009 (UTC)
