Simplify the following :
(i) `1+ i^(5)+i^(10)+i^(15)`
(ii) `(1+i)^(4)+(1+(1)/(i))^(4)`
(iii) `i^(n)+i^(n+1)+i^(n+2)+i^(n+3)`

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)

Simplify the following : (i) i^7 " " (ii) i^(1729) " " (iii) i^(-1924) + i^(2018) " " (iv) sum_(n=1)^(102) i^(n) " " (v) i i^2 i^3 …..i^(40)

1 + i^(2n) + i^(4n) + i^(6n)

Evaluate the following: (i) i^(135) " "(ii) i^(19) " " (iii) i^(-999) " " (iv) (-sqrt(-1))^(4n + 2), n in N .

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 .

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 Ndot