推 topee:感謝......... 04/07 22:19
※ 引述《topee (eason)》之銘言：
: find the principal value of
: a.
: e
: [---(-1-√3i)]^2πi
: 2
: b.
: (1-i)^4i
a. [(e/2)(-1-√3i)]^2πi = [(e/2)(2e^(4πi/3))]^2πi
= [e^(1+4πi/3)]^2πi = e^(2πi-8π^2/3)
= e^(-8π^2/3)e^(2πi) = e^(-8π^2/3)
b. (1-i)^4i = exp[Ln(1-i)^4i] = exp[4iLn(1-i)]
= exp[4i(ln√2-πi/4)] = exp[4iln√2+π]
= (e^π)(e^i4ln√2) = (e^π)(e^iln4)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.40.159.153