计算时，用到共轭，怎样才能让结果真正表示出来？

zhuozhwei

s = 2 ik/(E^(ik*11)*(ik*(Q[[1, 1]] - Q[[1, 2]]) + (Q[[2, 2]] - Q[[2, 1]])))

T = (s\[Conjugate])*s

为什么得到的计算结果是这样表示的------

(4 \[ExponentialE]^(-11 ik-11 Conjugate[ik]) ik \
Conjugate[ik])/(Conjugate[\[ImaginaryI] sin[\[ImaginaryI]] (-\
\[ImaginaryI] cos[3] sin[\[ImaginaryI]]-3 cos[\[ImaginaryI]] \
sin[3])^5+cos[\[ImaginaryI]] (cos[\[ImaginaryI]] cos[3]-1/3 \
\[ImaginaryI] sin[\[ImaginaryI]] sin[3])^5+ik (\[ImaginaryI] sin[\
\[ImaginaryI]] (-\[ImaginaryI] cos[3] sin[\[ImaginaryI]]+1/3 cos[\
\[ImaginaryI]] sin[3])^5+cos[\[ImaginaryI]] (cos[\[ImaginaryI]] \
cos[3]+3 \[ImaginaryI] sin[\[ImaginaryI]] sin[3])^5)] (\[ImaginaryI] \
sin[\[ImaginaryI]] (-\[ImaginaryI] cos[3] sin[\[ImaginaryI]]-3 cos[\
\[ImaginaryI]] sin[3])^5+cos[\[ImaginaryI]] (cos[\[ImaginaryI]] \
cos[3]-1/3 \[ImaginaryI] sin[\[ImaginaryI]] sin[3])^5+ik (\
\[ImaginaryI] sin[\[ImaginaryI]] (-\[ImaginaryI] cos[3] sin[\
\[ImaginaryI]]+1/3 cos[\[ImaginaryI]] sin[3])^5+cos[\[ImaginaryI]] \
(cos[\[ImaginaryI]] cos[3]+3 \[ImaginaryI] sin[\[ImaginaryI]] \
sin[3])^5)))

居然还带着Conjugate，很纳闷``望哪位大虾指点下，谢。

madio

是不是应该写成这个样子：

s = 2 ik/(E^(ik*11)*(ik*(Q[[1, 1]] - Q[[1, 2]]) + (Q[[2, 2]] -
Q[[2, 1]])))

T = (s/[Conjugate])*s