Evaluate :
(i) `i^(135)`
(ii) `i^((1)/(47)`
(iii) `(-sqrt(-1))^(4n +3) , n in N`
(iv) `sqrt(-25) + 3sqrt(-4) + 2 sqrt(-9)`

Explanation

(i) 135 leaves reaminder as 3 when it is divided by 4. Therefore,
`i^(135) = i^(3) = - i`
(ii) `i^((1)/(47)) = (1)/(i^(44)i^(3)) = (1)/(-i^(2))=i`
(iii) `(-sqrt(-1))^(4n + 3) = (-i)^(4n + 3)`
` =(-i)^(4n) (-i)^(3)`
`= {(-i)^(4n) } (-i)^(3)`
`= 1 xx (-i)^(3) = - i^(3) = -i^(3) = - (-i)=i`
(iv) `sqrt(-25) + 3sqrt(-4) + 2 sqrt(-9) = 5i + 6 i + 6i = 17i`