11/05/08 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),
11/05/08 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),