Algebra 2 Question... Plz help!?

Question

How do I simplify the complex number i^21 as much as possible ?

Answer 1

ixi = -1
ixixixi= 1
So i^21 = i^20 x i
= i
because i^20 = 1

Answer 2

i @@@@@@@@@@@@ i^5 = (i^4)(i) = (1)(i) = i @@ i^9 = i @@@@@ etc....

i^2 = -1 @@@@@@@@@ i^6 = (i^5)(i) = (i)(i) = -1 @@ i^10 = -1

i^3 = (i^2)(i) = (-1)(i) = -i @@ i^7 = (i^6)(i) = (-1)(i) = -i @ i^11 = -i

i^4 = (i^3)(i) = (-i)(i) = 1 @@ i^8 = (i^7)(i) = (-i)(i) = 1 @@ i^12 = 1

Starting with the first power of i the pattern i, -1, -i, 1 repeats for every four powers of i.

Divide the exponent of i by 4 and use the remainder to determine which number of the pattern is the value of that particular power of i. Remainder of 1, 2, 3, or 0 means the value is i, -1, -i, or 1, respectively.

i^7 7 divided by 4 gives remainder 3 So, i^7 = -i.

i^18 18 divided by 4 gives remainder 2 So, i^18 = -1.

i^32 32 divided by 4 gives remainder 0 So, i^32 = 1.

i^21 21 divided by 4 gives remainder 1 So, i^21=i

Answer 3

i^2 = - 1
i^4 = 1
i^21 = i(i^4)^5 = i(1)^5 = i

Answer 4

im so bad at math!!!