Question

求复数（1+i)^i的实部和虚部

Answer 1

(1+i)^i=e^(i*ln(1+i))

ln(1+i)=ln(1/√2+1/√2i)+ln(√2) = (π/4)i+1/2*ln(2)

i*ln(1+i) = -π/4 +1/2*ln(2) i

e^ (-π/4 +1/2*ln(2) i ) = e^(-π/4) * e^(1/2*ln(2) i )

= e^(-π/4) * ( cos(ln(2)/2) + i * sin(ln(2)/2) )

因此：

(1+i)^i 的实部 e^(-π/4) * cos(ln(2)/2) = 0.428829006

(1+i)^i 的虚部 e^(-π/4) * sin(ln(2)/2) = 0.154871752

o(∩_∩)o