數學問題 複數

1. 改為 x^2=1+i √3
Sol
x^2/2=(1/2)+i(√3/2)
=Cos(π/3)+iSin(π/3)
=Cos(2nπ+π/3)+iSin(2nπ+π/3)
x=√2[Cos(nπ+π/6)+iSin(nπ+π/6)]
(1) n=0
x=√2[Cos(π/6)+iSin(π/6)]
=√2[(√3/2 )+i(1/2)]
=√6/2 +i(√2/2)
(2) n=1
x=√2[Cos(7π/6)+iSin(7π/6)]
x=√2[Cos(7π/6)+iSin(7π/6)]
=√2[(－√3/2 )+i(－1/2)]
=－√6/2－i(√2/2)

2. x^3=－8i
Sol
x^3=8(Cosπ+iSinπ)=8[Cos(2nπ+π)+iSin(2nπ+π)]
x=2｛Cos[(2nπ+π)/3]+iSin[(2nπ+π)/3]｝
(1) n=0
x=2[Cos(π/3)+iSin[(π/3)]
=2[(1/2)+i(√3/2)]
=1+i√3
(2) n=1
x=2(Cosπ+iSinπ)=－2
(3) n=2
x=2[Cos(5π/3)+iSin(5π/3)]
=2[(1/2)－i√3/2]
=1－i√3

3. x^4=1
Sol
1=Cos0+iSin0=Cos(2nπ)+iSin(2nπ)
x= Cos(2nπ/4)+iSin(2nπ/4)= Cos(nπ/2)+iSin(nπ/2)
(1) n=0
x=Cos0+iSin0=1+0i=1
(2) n=1
x=Cos(π/2)+iSin(π/2)=i
(3) n=2
x= Cos(π)+iSin(π)=－1
(3) n=3
x= Cos(3π/2)+iSin(3π/2)=－i

第2 題

x^3 = -8i = 8(Cosπ+iSinπ) ??

似乎有誤 ! [ (3/2) (Pi) ]