Recent content by jrp131191

Yes sorry I forgot to mention that, I did write " given that C does not cross the origin". Edit: Ok I did mention it in the post lol. But I just want to know how I went with it. I was very scared that I wouldn't be able to do any of the proofs but I felt pretty confident about this one.

Hi, I just had my exam on complex analysis and would just like to know if I did this question correctly. It said that the function f(z) was analytic and to show that the integral of f(z)-\frac{c}{z} existed for some constant c, then to find a formula for c in term of an integral of f(z). I...

Hi I am doing a past exam for my complex analysis course and I should just mention right now that while it's a mix of pure/applied math I have never done a pure math unit before and i really really really suck at doing proofs and such.. Given c>0 and f(z) is entire such that |f(z)| ≤ c|z|...

Yup thanks, I've done it by looking over at my lecturers notes but I am very so confused as to something he had done. I was fine showing that the arc segments of the contour equal zero but when it came to the straight segments in the dog-bone I had to refer to his notes (He actually did the...

Nobody? :( It would help just to even know that I have come to the right final solution!

Homework Statement A semi-infinite bar (0 < x < 1) with unit thermal conductivity is fully insulated at x = 0, and is constantly heated at x = 1 over such a narrow interval that the heating may be represented by a delta function: \frac{\partial U}{\partial t}=\frac{\partial^2 U}{\partial...

Hi, first time trying to to use the complex inversion formula and I'm confusing myself. I'm trying to find L^-1{ ln(1 + (1/s))} I can see that there is an essential pole at s=0 (is this even right? The power series has an infinite amount of z^n's in the denominator..) and branch points at...

OK I FINALLY GOT IT, I'VE BEEN WORKING ON THIS FOR 2 DAYS! I essentially used your method which I did actually try in a much longer way yesterday. The problem was that my lecture notes say that you can manipulate \frac{1}{1+z^2} Into \frac{1}{1-(-z^2)} And thus \frac{1}{1+z^2} = \Big(...

Still ending up with two series that I cannot show are equal. I completely understand everything you have said but the only problem is that I need to use the power series of the logs!... I just essentially tried what you said but instead of directly differentiating (1/2i)Log((1+iz)/(1-iz)) I...

Oh no I do sorry, I think I misinterpreted what you were saying. My only problem is that the question explicitly tells me to obtain the power series for the logs about z=0 and use them to establish the identity. I will stare at what you wrote for a while and see if I can put that to use using...

Ok this will take me a while but I will show my working and what I CAN deduce.. \frac{1}{2i}log(\frac{1+iz}{1-iz}) = \frac{1}{2i}(\Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}(iz)^{n+1}}{n+1} \Big) + \Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}(-iz)^{n+1}}{n+1} \Big)) \frac{1}{2i}log(\frac{1+iz}{1-iz})...

Okay thanks, I do understand that but like I said, even as they are, I can't manipulate my two series to look identical already, so even if I differentiate them I will just end up with the same problem.. :(

I should not have posted this question here, I am trying to move it at the moment because it is homework. But yes I am asked to use series. I know how to derive it using the exponentials of sin & cos. I read that if I can show that for some c in U in which both my 2 series I am comparing are...

Hi, I am still trying to learn latex so this might fail but anyway, I am trying to establish the identity arctan(z) = \frac{1}{2i}log(\frac{1+iz}{1-iz}) By using the power series of log(1+iz) and log(1-iz) I have found the power series for both of these functions as well as arctan(z)...

Ok I think I figured it out.. I'm integrating an even function so if my integral is I(z) then I(z)=(1/2)J(z) where J(z) is the integral from -inf to inf... and I've actually integrated J(z) correct? I've just kind of read this off an example I've seen so If anyone can explain to me what was...