1.

Show that `(-sqrt(-1))^(4n+3) =i`, where n is a positive integer.

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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