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Show that `(-sqrt(-1))^(4n+3) =i`, where n is a positive integer. |
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Answer» We have `(-sqrt(-1))^(4n+3)=(-i)^(4n+3)" "[because sqrt(-1)=i]` `=(-i)^(4n)xx(-i)^(3)` `={(-i)^(4)}^(n) xx (-i^(3))=(1 xx i) = i" "[because (-i)^(4) = 1 and -i^(3) = -(-i) = i]` Hence, `(-sqrt(-1))^(4n+3)=i`. |
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