Nội dung bài giảng
Bài 33. Tính \({\left( {\sqrt 3 - i} \right)^6};\,\,\,{\left( {{i \over {1 + i}}} \right)^{2004}};\,\,\,{\left( {{{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }}} \right)^{21}}\)
Giải
\({\left( {\sqrt 3 - i} \right)^6} = {\left[ {2\left( {\cos \left( { - {\pi \over 6}} \right) + i\sin \left( { - {\pi \over 6}} \right)} \right)} \right]^6} = {2^6}\left[ {\cos \left( { - \pi } \right) + i\sin \left( { - \pi } \right)} \right] = - {2^6}\)
\({i \over {i + 1}} = {{1 + i} \over 2} = {1 \over {\sqrt 2 }}\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)\) nên
\(\eqalign{ & {\left( {{1 \over {1 + i}}} \right)^{2004}} = {1 \over {{2^{1002}}}}\left( {\cos {{2004\pi } \over 4} + i\sin {{2004\pi } \over 4}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{2^{1002}}}}\left( {\cos \pi + i\sin \pi } \right) = - {1 \over {{2^{1002}}}} \cr} \)
\({{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }} = {{\left( {5 + 3i\sqrt 3 } \right)\left( {1 + 2i\sqrt 3 } \right)} \over {1 + 12}} = {{ - 13 + 13i\sqrt 3 } \over {13}} = - 1 + i\sqrt 3 \)
\( = 2\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right) = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)\)
Do đó:
\({\left( {{{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }}} \right)^{21}} = {2^{21}}\left( {\cos 14\pi + i\sin 14\pi } \right) = {2^{21}}\)