\(\eqalign{ & a)\;\;z_1=\sqrt 2 \left( {\sqrt 3 - i} \right) = 2\sqrt 2 \left[ {\cos \left( { - {\pi \over 6}} \right) + i\sin \left( { - {\pi \over 6}} \right)} \right], \cr & {z_2} = 2\left( { - 1 - i} \right) = 2\sqrt 2 \left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right], \cr & {z_3} = {{{z_1}} \over {{z_2}}} = \cos \left( { - {\pi \over 6} + {{3\pi } \over 4}} \right) + i\sin \left( { - {\pi \over 6} + {{3\pi } \over 4}} \right) = \cos \left( {{{7\pi } \over {12}}} \right) + i\sin \left( {{{7\pi } \over {12}}} \right) \cr} \)
b) Mặt khác \({{{z_1}} \over {{z_2}}} = {{\sqrt 6 - i\sqrt 2 } \over { - 2 - 2i}} = {{\left( {\sqrt 6 - i\sqrt 2 } \right)\left( { - 2 + 2i} \right)} \over 8} = {{ - \sqrt 6 + \sqrt 2 } \over 4} + {{\sqrt 6 + \sqrt 2 } \over 4}i\) nên so sánh với kết quả câu a), suy ra:
\(\cos {{7\pi } \over {12}} = {{ - \sqrt 6 + \sqrt 2 } \over 4};\,\sin {{7\pi } \over {12}} = {{\sqrt 6 + \sqrt 2 } \over 4}\)
