0049132人目の素数さん垢版 | 大砲2011/05/08(日) 16:50:21.58 >>1 1+i = (√2)e^(iπ/4) = e^{ln(√2) +i(π/4)}, (1+i)^(1+i) = e^{[ln(√2) +i(π/4)](1+i)} = e^{ln(√2)-(π/4)}・e^{i[ln(√2)+(π/4)]} =(√2)e^(-π/4)・e^(i[ln(√2)+(π/4)]} = r・e^(iθ), ここに r = (√2)e^(-π/4), θ = (π/4) + ln(√2),