a) Ta có:
\(\eqalign{
& {{4i} \over {1 + i\sqrt 3 }} = {{4i(1 - i\sqrt 3 )} \over 4} = \sqrt 3 + i \cr
& = 2({{\sqrt 3 } \over 2} + {1 \over 2}i) = 2(cos{\pi \over 6} + {\rm{i}}\sin {\pi \over 6}) \cr} \)
Suy ra: \({({{4i} \over {1 + i\sqrt 3 }})^6} = {2^6}(cos\pi + \,i\sin \pi ) = - {2^6}\)
b) Ta có:
\(\eqalign{
& {(\sqrt 3 + i)^5} = {2^5}(cos{{5\pi } \over 6} + {\rm{i}}\sin {{5\pi } \over 6})\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \cr
& 1 - i\sqrt 3 = 2({1 \over 2} - {{\sqrt 3 } \over 2}i) \cr
& = 2(cos( - {\pi \over 3}) + {\rm{i}}\sin ( - {\pi \over 3})) \cr
& \Rightarrow {(1 - i\sqrt 3 )^{11}} = {2^{11}}{\rm{[cos(}}{{ - 11\pi } \over 3}) + {\rm{isin(}}{{ - 11\pi } \over 3}){\rm{]}}\,\,\,\,\,\,\,\,\,(2) \cr} \)
Từ (1) và (2) suy ra:
\(\eqalign{
& {{{{(\sqrt 3 + i)}^5}} \over {{{(1 - i\sqrt 3 )}^{11}}}} = {1 \over {{2^6}}}{\rm{[cos(}}{{5\pi } \over 6} + {{11\pi } \over 3}) + {\rm{i}}\sin {\rm{(}}{{5\pi } \over 6} + {{11\pi } \over 3}){\rm{]}} \cr
& = {1 \over {{2^6}}}(cos{{9\pi } \over 2} + {\rm{i}}\sin {{9\pi } \over 2}) = {i \over {64}} \cr} \)
