（2007•茂名二模）定义：eiθ=cosθ+isinθ（i为虚数单位），若ei2π3+1-3i=eiα，则α角可能是（）A．4π3B．7π6C．5π3D．11π6

Question

（2007•茂名二模）定义：e^iθ=cosθ+isinθ（i为虚数单位），若ei(2π)/(3)+1-√(3)i=e^iα，则α角可能是（　　）

A．(4π)/(3)
B．(7π)/(6)
C．(5π)/(3)
D．(11π)/(6)

Answer 1

由e^iθ=cosθ+isinθ，得
e^iα=cosα+isinα，
∴ei(2π)/(3)+1-√(3)i=cos(2π)/(3)+isin(2π)/(3)+1−√(3)i
=−(1)/(2)+(√(3))/(2)i+1−√(3)i=(1)/(2)−(√(3))/(2)i=cos(5π)/(3)+isin(5π)/(3)=ei(5π)/(3)=e^iα．
∴α＝(5π)/(3)．
故选：C．