Question:

How do I simplify: i^7, i^28, i^76?

Answers:

A-Best: The important things to remember here are (1) i^4 = 1 , and (2) i^(r + q * m) = (i^r) * ((i^m)^q) . Say we want to simplify i^n for some positive integer n. Using the division algorithm, we may find numbers q and r (the quotient and remainder) where 0 ≤ r < 4 such that n = r + 4 * q . Thus i^n = i^(r + 4 * q) = (i^r) * (i^4)^q = i^r . In other words, we just need to look at the remainder left after division by 4, and compute i raised to that remainder. The remainder will be one of 0, 1, 2, or 3, and so we just need to know that i^0 = 1, i^1 = i, i^2 = -1, and i^3 = -i . We have 7 = 3 + 4 * 1 , 28 = 0 + 4 * 7 , and 76 = 0 + 4 * 19 , and so compute that i^7 = i^3 = -i and i^(28) = i^(76) = 1 .  

A: i = i i^2 = -1 i^3 = -i i^4 = 1 then pattern repeats so i^7 = i^3 = -i i^28 = i^4 = 1 i^76 = i^4 = 1 (28 and 76 are both multiples of 4)  

A: Powers of i are a repeating pattern: i¹=i i²=-1 i³=-i i⁴=1 i⁵ is just i⁴*i, and since i⁴=1, i⁵=(1)i, or simply, i. You can use the same concept for i⁷, i²⁸, and i⁷⁶ i⁷=(i⁴)(i³)=(1)i³=i³=-i i²⁸=(i⁴)⁷=(1)⁷=1 i⁷⁶=(i⁴)¹⁹=(1)¹⁹=1 Hope that helps!  

A: i^7=i^4.i^3 =1.i^3 =-i because i^4=1,i^3=-i i^28=(i^4)^7=1^7=1 i^76=(i^4)^19=1  

A: i^2 = -1 i^3 = i*i^2 = -i i4 = i^2 * i^2 = 1 i^7 = i^3 * i ^4 = -i * 1 = -i i^28 = i^4*i^4*i^4*i^4*i^4*i^4*i^4 = 1 i^76 = i^28*i^28*i^4*i^4*i^4*i^4*i^4 = (1)(1) = 1  

A: the definition of i is the square root of -1, so i^2 = -1, i^3 = i*i^2 = -i, i^4 = i^2*i^2 = 1 So i^7=i^3*i^4 = -1 i^28 = i^(4*7) = 1 i^76 - i^(4*19) = 1  

A: i² = -1 i³ = -i i^4 = 1 i^7 = (i^4)(i^3) = (1)(-i) = -i i^28 = (i^4)^7 = (1)^7 = 1 i^76 = (i^4)^19 = (1)^19 = 1 Answer: i^7 = -i, i^28 = 1, and i^76 = 1