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Find-the-value-of-i-n-for-every-positive-integer-n-where-i-2-1-i-3-i-2-i-i-4-i-2-i-2-etc-




Question Number 190275 by aye786naing last updated on 30/Mar/23
 Find the value of i^n  for every positive   integer n, where i^2  = −1, i^3 = i^2 i, i^4  = i^2 i^2  , etc.
$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{i}^{\mathrm{n}} \:\mathrm{for}\:\mathrm{every}\:\mathrm{positive} \\ $$$$\:\mathrm{integer}\:\mathrm{n},\:\mathrm{where}\:\mathrm{i}^{\mathrm{2}} \:=\:−\mathrm{1},\:\mathrm{i}^{\mathrm{3}} =\:\mathrm{i}^{\mathrm{2}} \mathrm{i},\:\mathrm{i}^{\mathrm{4}} \:=\:\mathrm{i}^{\mathrm{2}} \mathrm{i}^{\mathrm{2}} \:,\:{etc}. \\ $$
Answered by Frix last updated on 31/Mar/23
k∈N: i^n = { ((1; n=4k)),((i; n=4k+1)),((−1; n=4k+2)),((−i; n=4k+3)) :}
$${k}\in\mathbb{N}:\:\mathrm{i}^{{n}} =\begin{cases}{\mathrm{1};\:{n}=\mathrm{4}{k}}\\{\mathrm{i};\:{n}=\mathrm{4}{k}+\mathrm{1}}\\{−\mathrm{1};\:{n}=\mathrm{4}{k}+\mathrm{2}}\\{−\mathrm{i};\:{n}=\mathrm{4}{k}+\mathrm{3}}\end{cases} \\ $$

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