If z=eiθz = e^{i\theta}z=eiθ, which of the following is equivalent to sinθ\sin\thetasinθ?
z+zˉ2\frac{z + \bar{z}}{2}2z+zˉ
z−zˉ2i\frac{z - \bar{z}}{2i}2iz−zˉ
z−zˉ2\frac{z - \bar{z}}{2}2z−zˉ
z+zˉ2i\frac{z + \bar{z}}{2i}2iz+zˉ