=-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))的指数表达式为（ ）。A =-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))B =-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))C =-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))D =-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))

步骤 1：将复数转换为标准形式
$z=-\sqrt {2}(\sin \dfrac {\pi }{6}+i\cos \dfrac {\pi }{6})$
根据三角函数的性质，$\sin \dfrac {\pi }{6}=\cos \dfrac {\pi }{3}$，$\cos \dfrac {\pi }{6}=\sin \dfrac {\pi }{3}$，所以
$z=-\sqrt {2}(\cos \dfrac {\pi }{3}+i\sin \dfrac {\pi }{3})$
步骤 2：将复数转换为指数形式
复数的指数形式为$a+bi=r(\cos \theta +i\sin \theta )=r{e}^{i\theta }$
所以，$z=-\sqrt {2}(\cos \dfrac {\pi }{3}+i\sin \dfrac {\pi }{3})=-\sqrt {2}{e}^{\dfrac {\pi }{3}}$