Bài 33 trang 207 SGK giải tích 12 nâng cao, Tính...

Bài 33. Tính ({left( {sqrt 3  – i} ight)^6};,,,{left( {{i over {1 + i}}} ight)^{2004}};,,,{left( {{{5 + 3isqrt 3 } over {1 – 2isqrt 3 }}} ight)^{21}})

Giải

({left( {sqrt 3  – i} ight)^6} = {left[ {2left( {cos left( { – {pi  over 6}} ight) + isin left( { – {pi  over 6}} ight)} ight)} ight]^6} = {2^6}left[ {cos left( { – pi } ight) + isin left( { – pi } ight)} ight] =  – {2^6})

({i over {i + 1}} = {{1 + i} over 2} = {1 over {sqrt 2 }}left( {cos {pi  over 4} + isin {pi  over 4}} ight)) nên

(eqalign{  & {left( {{1 over {1 + i}}} ight)^{2004}} = {1 over {{2^{1002}}}}left( {cos {{2004pi } over 4} + isin {{2004pi } over 4}} ight)  cr  & ,,,,,,,,,,,,,,,,,,,,,, = {1 over {{2^{1002}}}}left( {cos pi  + isin pi } ight) =  – {1 over {{2^{1002}}}} cr} )

({{5 + 3isqrt 3 } over {1 – 2isqrt 3 }} = {{left( {5 + 3isqrt 3 } ight)left( {1 + 2isqrt 3 } ight)} over {1 + 12}} = {{ – 13 + 13isqrt 3 } over {13}} =  – 1 + isqrt 3 )

( = 2left( { – {1 over 2} + {{sqrt 3 } over 2}i} ight) = 2left( {cos {{2pi } over 3} + isin {{2pi } over 3}} ight))

Do đó:

({left( {{{5 + 3isqrt 3 } over {1 – 2isqrt 3 }}} ight)^{21}} = {2^{21}}left( {cos 14pi  + isin 14pi } ight) = {2^{21}})