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

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

Evaluate : (i^(18)+((1)/(i))^(25))

Evaluate [i^(18)+(1/i)^(25)]^(2) .

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

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

Prove that: (i) (1-i)^(2)=-2i (ii) (1+i)^(4)xx(1+(1)/(i))^(4)=16 (iii) {i^(19)+((1)/(i))^(25)}^(2)=-4 (iv) i^(4n)+i^(4n+1)+i^(4n+2)+i^(4n+3)=0 (v) 2i^(2)+6i^(3)+3i^(16)-6i^(19)+4i^(25)=1+4i .

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 N.

Express i^(18) + ((1)/(i))^(25) in a + ib form.

Show that : (i) {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 N .

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