解:
e^(it)=cost+isint
据此可知:
(1+i)^i
=[e^(ln(1+i))]^i
=e^(i*ln(1+i))
=e^[i*ln(2^(1/2)(cosPi/4+i*sinPi/4))]
=e^[i*(ln2/2+i*Pi/4)] 因为e^(iPi/4)=cosPi/4+isinPi/4 所以:ln(cosPi/4+isinPi/4)=iPi/4
=e^(-Pi/4+iln2/2)
=e^(Pi/4)^(-1)(cos(ln2/2)+isin(ln2/2))