line integral

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

法3

自創路徑

令C：re^iθ

r→0+

∮phi( 1／z )dz = ∮c( 1／z )dz

this is closed line integral.

the answer is 2πi