express i^n + i^(n+1) + i^(n+2) in terms of a +ib
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i^n + i^(n+1) + i^(n+2)
= iⁿ + i¹.iⁿ + i².iⁿ
= iⁿ ( 1 + i¹ + i²)
we know, i² = -1 use this above
=iⁿ( 1 + i -1)
= iⁿ.i
= i^(n+1)
use De - Moivre's theorem ,
i^(n+1) = (cosπ/2 + isinπ/2)^(n+1)
= cos(n +1)π/2 + isin(n+1)π/2
hence, cos(n+1)π/2 + isin(n+1)π/2 is in the form of a + ib
= iⁿ + i¹.iⁿ + i².iⁿ
= iⁿ ( 1 + i¹ + i²)
we know, i² = -1 use this above
=iⁿ( 1 + i -1)
= iⁿ.i
= i^(n+1)
use De - Moivre's theorem ,
i^(n+1) = (cosπ/2 + isinπ/2)^(n+1)
= cos(n +1)π/2 + isin(n+1)π/2
hence, cos(n+1)π/2 + isin(n+1)π/2 is in the form of a + ib
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