Bài 33 trang 207 SGK giải tích 12 nâng cao

Đề bài

Tính ${\left( {\sqrt 3  - i} \right)^6};{\left( {{i \over {1 + i}}} \right)^{2004}};{\left( {{{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }}} \right)^{21}}$

Lời giải chi tiết

$\begin{array}{l} \sqrt 3 - i = 2\left( {\frac{{\sqrt 3 }}{2} - \frac{1}{2}i} \right)\\ = 2\left( {\cos \left( { - \frac{\pi }{6}} \right) + i\sin \left( { - \frac{\pi }{6}} \right)} \right) \end{array}$

$\Rightarrow {\left( {\sqrt 3  - i} \right)^6}$ $= {\left[ {2\left( {\cos \left( { - {\pi  \over 6}} \right) + i\sin \left( { - {\pi  \over 6}} \right)} \right)} \right]^6}$ $= {2^6}\left[ {\cos \left( { - \pi } \right) + i\sin \left( { - \pi } \right)} \right] =  - {2^6}$

${i \over {i + 1}} = \frac{{i\left( {1 - i} \right)}}{{1 + 1}} = {{1 + i} \over 2}$ $= \frac{1}{2}\left( {1 + i} \right) = \frac{1}{2}.\sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}i} \right)$ $= {1 \over {\sqrt 2 }}\left( {\cos {\pi  \over 4} + i\sin {\pi  \over 4}} \right)$

$\eqalign{  &  \Rightarrow{\left( {{i \over {1 + i}}} \right)^{2004}} \cr &= {1 \over {{{(\sqrt 2)}^{2004}}}}\left( {\cos {{2004\pi } \over 4} + i\sin {{2004\pi } \over 4}} \right)  \cr  &  = {1 \over {{2^{1002}}}}\left( {\cos \pi  + i\sin \pi } \right) =  - {1 \over {{2^{1002}}}} \cr}$

${{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }} = {{\left( {5 + 3i\sqrt 3 } \right)\left( {1 + 2i\sqrt 3 } \right)} \over {1 + 12}}$ $= {{ - 13 + 13i\sqrt 3 } \over {13}} =  - 1 + i\sqrt 3$

$= 2\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right)$ $= 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$

Do đó:

${\left( {{{5 + 3i\sqrt 3 } \over {1 - 2i\sqrt 3 }}} \right)^{21}} = {2^{21}}\left( {\cos 14\pi  + i\sin 14\pi } \right)$ $= {2^{21}}$

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