Can -1^ (1/4) be expressed in terms of i Solution To express (-1)^1/4 and (-1)^(1/6) interms of i. We know that i= (-1)^(1/2) or i^2 = -1. -1 = cos pi +sinpi = cos(2n+1)pi +isin(2n+1)pi Therefore (-1)^1/4 = Cos(2n+1)pi/4+sin(2n+1)pi)/4 = cos(2n+1)pi/2+isin(2n+1)pi/2 , for n = 0,1,2... 1/2^(1/2) +i/sqrt2 , -1/2^(1/2) +i/2^1/2) , -1/2^1/2) -i/2^(1/2) and (1/2^(1/2) +i/2^(1/2). (-1)^1/6 = {cos(2npi+1) +isin(2npi+1)}^(1/6) = cos(2n+1)pi/6 +isin(2n+1)pi/6. for n = 0,1,2,3,.. = (1/2)*3^(1/2) +i(1/2) , i , -(1/2)3^(1/2 ) +i/2 , (1/2)*3^(1/2)-i/2, -i , (1/2)3^(1/2) - i ,.

1. Can -1^ (1/4) be expressed in terms of i
Solution
To express (-1)^1/4 and (-1)^(1/6) interms of i.
We know that i= (-1)^(1/2) or i^2 = -1.
-1 = cos pi +sinpi = cos(2n+1)pi +isin(2n+1)pi
Therefore (-1)^1/4 = Cos(2n+1)pi/4+sin(2n+1)pi)/4 = cos(2n+1)pi/2+isin(2n+1)pi/2 , for n =
0,1,2...
1/2^(1/2) +i/sqrt2 , -1/2^(1/2) +i/2^1/2) , -1/2^1/2) -i/2^(1/2) and (1/2^(1/2) +i/2^(1/2).
(-1)^1/6 = {cos(2npi+1) +isin(2npi+1)}^(1/6)
= cos(2n+1)pi/6 +isin(2n+1)pi/6. for n = 0,1,2,3,..
= (1/2)*3^(1/2) +i(1/2) , i , -(1/2)3^(1/2 ) +i/2 , (1/2)*3^(1/2)-i/2, -i , (1/2)3^(1/2) - i ,