Evaluate: `[i^19+(1/i)^25]^2`

Evaluate : [i^(18)+(1/i)^(25)]^3

Evaluate: [i^(18) + ((1)/(i))^(25)]^(3)

Evaluate [ i^(14) - (1/i)^(34)]^2

Evaluate (i) ┌1 (ii) ┌2

Show that: {i^(19)+(1/i)^(25)}^2=-4 (ii) {i^(17)-(1/i)^(34)}^2=2i (iii) {i^(18)+(1/i)^(24)}^3=0 (iv) i^n+i^(n+1)+i^(n+2)+i^(n+3)=0 for all n in Ndot

Simplify the following : (i) [i^(19) +(1)/(i^(25))]^(2) (ii) [i^(5)- (1)/(i^(3))]^(4)

Simplify: [i^(18) + ((1)/(i))^(25)]^(3)

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

Evaluate : (i) (a+1) (a-1) (a^2 + 1)

Evaluate 2i^(2)+ 6i^(3)+3i^(16) -6i^(19) + 4i^(25)