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B
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C
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D
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Solution
The correct option is C Given equation is (−12+i√32)365=4+5(cos2π3+isin2π3)334+3(cos2π3+isin2π3)365=4+5[cos6683π+isin6683π]3[cos7303π+isin7303π] =4+5[cos(222π+2π3)+isin(222π+2π3)]+3[cos(243π+π3)isin(243π+π3)] =4+5(cos2π3+isin2π3)+3(−cosπ3−isinπ3)=4+5(−12+i√32)+3(−12−i√32)=4−4+2i√32=i√3.