line integral

2010-02-08 1:43 pm
let phi(t) = acost +ibsint 0<=t<=2pi

evaluate [ integal of (1/z) dz ] in two ways
show this = 2pi / ab

thx
更新1:

let phi(t) = acost +ibsint 0 < = t < =2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab

更新2:

the integral is calculate at the region of phi

回答 (3)

2010-02-09 1:24 am
✔ 最佳答案
Method1: integrate directly
(1/z)dz=[(xdx+ydy)+i(xdy-ydx)]/(x^2+y^2)
=[ (-a^2 cost sint+b^2 sint cost)+i(ab)]dt/[(a cost)^2+(b sint)^2]
line integral= ∫[0~2π] (b^2-a^2)sint cost/[(acost)^2+(bsint)^2] dt
+iab∫[0~2π] 1/[(acost)^2+(bsint)^2] dt
= ln| (acost)^2+(bsint)^2 | sub. t=0~2π
+2iab∫[-π/2~π/2] (sect)^2/[a^2+(btant)^2] dt
= 0 + 2iab/(ab)*arctan(btant /a) for t=-π/2 ~ π/2
= 0 + 2i*(π/2 + π/2)= 2πi

Method2: by Residue thm.
line integral∫ dz/z
=2πi*(esidue of 1/z at the pole z=0)
= 2πi * 1= 2πi

2010-02-09 7:16 am
法3

自創路徑

令C:re^iθ

r→0+

∮phi( 1/z )dz = ∮c( 1/z )dz
2010-02-08 3:11 pm
this is closed line integral.

the answer is 2πi


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