Question

(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8

√3是根号3

Answer 1

-√3/2-1/2i=cos(-2π/3)+isin(-2π/3)
(2+2i)/(1-√3i)=2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]
所以(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8
=[cos(-2π/3)+isin(-2π/3)]^12+[2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]]^8
=e^-8πi+16e^14πi/3=1+16e^2πi/3=-8√3+16+8i

Answer 2

(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8
解答：
-√3/2-1/2i=cos(-2π/3)+isin(-2π/3)
(2+2i)/(1-√3i)=2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]
所以(-√3/2-1/2i)^12+[ (2+2i)/(1-√3i) ]^8
=[cos(-2π/3)+isin(-2π/3)]^12+[2√2(cosπ/4+isinπ/4)/2[cos(-π/3)+isin(-π/3)]]^8
=e^-8πi+16e^14πi/3=1+16e^2πi/3=-8√3+16+8i