Recent content by FelixHelix

OK... So: Let w = cot(z) and in exponential form i* \frac{\exp{(iz)}+\exp{(-iz)}}{\exp{(iz)}-\exp{(-iz)}} Then let y = exp(iz) so you get: y = i* \frac{w+w^{-1}}{w-w^{-1}} When I rearrange the equation for y I then solve a quadratic in w so: So the two answers to w are \pm...

Hi - Thanks for the post. I believe I am to consider the complex domain as well. My answersheet gives the cut to be (-i,i) but I don't see how you get there or limit it to be multivalued... All examples we've covered have solved for y after some re-expressing hence why I thought that might be...

Just covered branch cuts in my undergraduate course but stuck on one of the questions... Find the domain on which f(z) = arccot(z) is single valued and analytic. Now, we've looked at ln(z) in class and I understand the principal of limiting the domain but I'm not having much success and...

Ahhh, I see now. Double angle identity and then using Osbournes Rule... easy when you see it! Thanks vanhees71 for taking the time to show the working - It's much appreciated... FH

I suppose I mean the second transformation. I can't see how you get there... What identities are at work...?

Hi there. I've been trying to solve the integral of 1/(1+cosh(x)). I use Wolfram to give me a detailed solution but I still don't understand second transformations they use. I've attached a a screen grab of the workings and hoped someone could run through it with me. I've used the tan x =...

Ill take a look in the morning - thanks for your help!

yes a typo indeed - apologies. What substitution do u use to get there though - i don't see it!

?? I'm not sure??

I see,thats a good trick i hadnt thought of using. But how would you manipulate the integral you end up with?

No, the integrand is 1/(1+cosh(x))

I cannot reach the answer for this integral which is part of a bigger question related to discounting investments. I know what the answer to the integral is and have tried all the substitutions and tricks I know. Any pointer would be great! ∫(1/(1+cosh(x))) = tanh(x) + C Thanks, Felix

Hi Shredder - No I didn't know how to take the sqrt of a complex number... but I do now. Thanks for pointing this out - I looked it up and am happy with dealing with these now. Your help is much appreciated! Felix

I can: z1,z2= \frac{(2i+4) \pm \sqrt{(2i+4)^2-(4)(1)(8i)}}{(2)(1)} this simplifies to: z1,z2 = (i+2) \pm \sqrt{3-4i} Which isn't what I need... Do you get the solutions z = 4 and z = 2i?

Not sure if this is the right place to post (but its related to a complex analysis questions) I'm doing a past paper for my revision and am stuck at the first hurdle. I simply cannot factor this polynomial in z for the life of me. I've tried completing the square and the usual quadratic...