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Start by calculating i^1, i^2, i^3, and i^4. Then calculate i^5, i^6, i^7, and i^8. Do you notice any patterns?
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i^n= i, ifR=1 - 1, ifR=2 - i, ifR=3 1, ifR=0 Now, let's consider i^(41). We need to find the remainder when 41 is divided by 4. 41 Ă· 4 = 10 R1 Because the remainder is 1, we get that i^(41)=i.
The Commutative and Associative Properties of Multiplication hold true for imaginary numbers. We can use them to find the powers of i. Let's consider the first four powers. Recall that any number raised to the power of one equals itself, so i^1=i. Moreover, by definition i=sqrt(- 1), so we know that i^2= - 1. ccccccc i^1&=& i i^2&=& - 1 i^3&=& i^2 * i &=& - 1 * i&=& - i i^4&=& i^2 * i^2&=& ( - 1)^2 * ( - 1)^2&=& 1 Let's now calculate the following four powers. To do so, we will use the results obtained above. ccccccc i^5 &=& i^4 * i &=& 1 * i &=& i i^6 &=& i^4 * i^2 &=& 1 * ( - 1) &=& - 1 i^7 &=& i^4 * i^3 &=& 1 * ( - i) &=& - i i^8 &=& i^4 * i^4 &=& 1 * 1 &=& 1 Notice that the pattern i, - 1, - i, 1, ... repeats in that order continuously after the first four results.