If (root(3)+i)^10=a+bi then a and b are

Dear Student,
Please find below the solution to the asked query:

We have3+i10=a+ibNowz=3+iz=32+12=2α=tan-113=π6Hence 3+i=2cosπ6+isinπ6Hence 3+i10=a+ib becomes2cosπ6+isinπ610=a+ib210cosπ6+isinπ610=a+ibUsing De-Movier's theorem210cos10π6+isin10π6=a+ib210cos5π3+isin5π3=a+ib210cos2π-π3+isin2π-π3=a+ib210cos2π-π3+isin2π-π3=a+ib210cosπ3-isinπ3=a+ib21012-i32=a+ib29-i293=a+ibComparinga=29b=293

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