Recent content by Smin0

Thank you! Oh my goodness, I see where it happened now, it was the multiplying out of the i(1 - sin^{2}\theta)cos\theta, I just ignored the -isin^{2}\theta\cos\theta term (and the similar (1 - cos^{2}) bit). I spent forty minutes yesterday on this really simple check of the Cauchy-Riemann...

Yep, that's the line integral. And I got the other part, from -1 to 1 along the x-axis, I just needed to learn to do the other part explicitly to prove that the closed contour sums to 0. I still can't see what's wrong with the integration in cosines and sines, but it comes out as the wrong...

Minus signs ... That should be -- 1 -- 1 -- 1 -- 1 = + 1 + 1 + 1 + 1 = 4 sorry. Still wrong though :(. Oh, and: should be \int{z}^{2}{dz} = \left[- cos\theta - cos^{3}\theta\right]^{\pi}_{0} + i\left[sin\theta - sin^{3}\theta\right]^{\pi}_{0}

o___o ... okay, and when I do it: \int{z}^{2}{dz} = \int^{\pi}_{0}(e^{i\theta})^{2}(ie^{i\theta}d\theta) \int{z}^{2}{dz} = \int^{\pi}_{0}(e^{i2\theta})(ie^{i\theta}d\theta) \int{z}^{2}{dz} = i\int^{\pi}_{0}(e^{i3\theta})d\theta \int{z}^{2}{dz} =...

Int((z^2)dz). Also: my GOODNESS that latex took a long time to type out. Okay, well I thought I'd found the trouble when I realized that dz = izd\theta instead of ird\theta, but alas it now comes out to -4. My working this time is as follows (r = 1): \int{z}^{2}{dz} =...

Help! I'm preparing for Prelims (exams taken at the end of the first year of my course in Physics) and I can't do the line integral of (z^2)dz around the open semi-circle counter-clockwise from x=1 to x=-1. However I try to tackle it the answer comes out as 0, when I know it has some size...

Thanks :)! Thanks, that worked out really well :). I was wondering though, how would one work it out from the form \int \frac{1}{u\sqrt{1-u^2}} du ? Could anyone help me see how to integrate it from this?

Help please I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.