Recent content by KNUCK7ES

thanks btw

Homework Statement I =\int \frac{cosx}{x^{2}-2x+2}dx the integral runs from -inf to inf evaluate the integral using the calculus of residues. Homework Equations shown in my attempt The Attempt at a Solution Re \oint\frac{e^{iz}}{z^{2}-2z+2} with singularities at...

ok i did so, so it the contour breaks up into for parts, Jo + JI + Jtop+ Jbottom where Jo means the outer part of the contour and JI is the inner part to the contour. Jo=JI=0. now i just do the calculus of residues on the Jtop and Jbottom, whic are \int\frac{1}{x\sqrt{x-2}} dx and...

Homework Statement compute I=∫_2^∞ (1/(x(x-2)^.5)) dx using the calculus of residues. be sure to choose an appropriate contour and to explain what happens on each part of that contour. Homework Equations transform to a complex integral I= ∮ (1/(z(z-2)^.5)) dz The Attempt...

actually upon further thinking of my question, it would have to have bounds ∮ 1/(z^2-z+2) dz for lzl=1 would be 0 but if i were to extend the radius out to lzl =2 then it gets more interesting

i guess my confusion is whether we can set f(z) = z/(9-z^2) OR f(z)= z/(2z-i). but i'd much rather just do z/(2z-i). so i suppose my question is just inquisitive. but i could see it expanded out to something like ∮ 1/(z^2-z+2) dz for lzl=1. in this case it isn't so easy to find an f(x) to...

i did both, and the partial fractions is a little more ugly. but where i became stuck, is that i began to wonder if i had to do it twice the other one being where f(z) = z/(2z-i)

so being aware of the poles, in the case of (b) we have poles at z=-3,3,i/2. but the only one that matters is the i/2 because it is the only one inside the contour line. my only problem is now trying to get ∮ z/((9-z^2)(2z-i)) dz into the form f(w) ∮ 1/(z-w) dz

kk thanks alot

so that would mean that the integrand is analytic in the interior and on the contour, so it would equal 0?

1. evaluate each integral around the counter-clockwise circle lzl =1 repeat for lzl=2. (a) ∮ cos(z)/(3z-pi) dz, (b) ∮ z /((9-z^2)*(2z-i)) dz thats all it says. i think that the pole may be not be significant

i can't believe i didn't factor. now i get 1/3*pi*i

Homework Statement i have the closed ∫ (cos(z))/(3z-π)) dz Homework Equations i presume this works closed ∫ f(z)/(z-w) dz = f(w)2πi The Attempt at a Solution i think the answer is cos(pi)*2πi