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
let phi(t) = acost +ibsint 0 < = t < =2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab
the integral is calculate at the region of phi
收錄日期: 2021-04-30 14:19:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100208000051KK00247