If $${e^{i\theta }} = \cos \theta + i\sin \theta $$    then for the $$\vartriangle ABC,{e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}}$$     is

Solution :
$$\eqalign{ & {e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}} = \left( {\cos A + i\sin A} \right)\left( {\cos B + i\sin B} \right)\left( {\cos C + i\sin C} \right) \cr & {e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}} = \cos \left( {A + B + C} \right) + i\sin \left( {A + B + C} \right) = \cos \pi + i\sin \pi . \cr} $$