求算法:極式問題

Sol(√3－i)^10=2^10*(√3/2－i/2)^10=1024*[Cos(330度)+iSin(330度)]^10=1024*[Cos(3300度)+iSin(3300度)]=1024*[Cos(3300度－9*360度)+iSin(3300度－9*360度)]=1024*[Cos(60度)+iSin(60度)]=1024*[(1/2)+i(√3/2)]=512+512√3i

本題用負角可讓計算量少些
(√3-i)^10

=2^10*(√3/2-i/2)^10

=2^10*[cos(-30度)+isin(-30度)]^10

=2^10*[cos(-300度)+isin(-300度)]

=2^10*[cos(60度)+isin(60度)]

=2^10*[(1/2)+i(√3/2)]

=512+512√3i