Question:

How do you solve the imaginary number i^22?
Solve the imaginary number i to the 22nd power. How do you do it? i^22

Answers:

A-Best: i^22 = i^(5(4) + 2) ....... = i^(20 + 2) ....... = i²⁰ i² ....... = (i⁴)⁵(i²) ....... = 1⁵(–1) ....... = –1  

A: i^2 = -1 i^22 = (i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2) Because you have to add the powers. so i^22 = (-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1) = -1  

A: i^2 = -1 i^22 = (i^2)^11 = -1^11 = -1  

A: i^2=-1 i^4=(-1)^2=1 i^20=(-1)^10=1^5=1. i^22=i^20*i^2 =1*(-1) = -1  

A: Write out the 4 possibilities. i = √-1 i^2 = (√-1)(√-1) = -1 i^3 = (√-1)(√-1)(√-1) = -1(√-1) = -√-1 = -i i^4 = (√-1)(√-1)(√-1)(√-1) = -1*-1 =1 this keeps repeating i^5 is the same as i i^6 is the same as i^2 To find i^22, divide 22 by 4. This gives us 5 with a remainder of 2. We only care about the remainder. Since it's 2, the answer would be i^2 which is -1. The rule is: 1. Divide by 4 and get the remainder. 2. The remainder is the power of i. (if there is no remainder then the power is 4 because 4 goes into the number perfectly) Examples: i^29 --> 29/4 = 7, remainder 1 ---> i^1 = i i^300 --->300/4 = 75, no remainder ----> i^4 = 1 i^107 ----> 107/4 = 26, remainder 3 ----> i^3 = -i That's all there is to it. Good Luck!