# 1. What pattern does the power of i follow?2. How can you simplify i^5689? What can be the base of your reasoning?3. What can support your reasoning?

###### Question:

2. How can you simplify i^5689? What can be the base of your reasoning?

3. What can support your reasoning?

## Answers

Answer:Step-by-step explanation:Start by doing a much simpler problemOnei^0 = 1i^1 = ii^2 = i^2 = - 1i^3 = i^3 = (-1) * i = -ii^4 = i^4 = (i)^2 * i^2 = -1 * -1 = 1So now you are ready to tackle the question you were given.TwoEvery 4th power of i repeats itself. You must get one of the four answers above. So divide 5689 by 4. What do you get?1422.25What does that mean?It means that you run through the cycle of four 1422 times. What does that mean?1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 ....................................*1There are 1422 ones in that expression. What about the 0.25? What about that?That means that you have one i left over. So the answer is iThreeThe simple question defines the pattern which will repeat itself over and over again. By knowing how the simple pattern works you can develop any power of i. For examplei^15622 = 3905.5 0.5 is the same as 2/4 i^2 = -1i^15622 = - 1

Answer:1. [tex]i = \sqrt{ - 1} \\ - 1 = {i}^{2} \\ - i = {i}^{3} \\ 1 = {i}^{4} (every \: multiple \: of \: four)[/tex]2.[tex]i = {i}^{5689} [/tex]Step-by-step explanation:3. After i⁴, the cycle repeats itself from the very beginning [i = i⁵ (and so on)]. Every 4⁄4 that go by, you start from the beginning, but you keep moving up one in the exponents.2. Since ¼ of 5689 is 1422¼, I know that it is imaginary, or iota, so moving one down would give 5688, where ¼ of that is 1422, with NOTHING left over, and that would equal to 1. You see, you have to know your multiples and divisiblity rules. The divisiblity rule of 4 states that if the LAST TWO DIGITS of an entire number are divisible by 4, then the ENTIRE number is divisible by 4. So, from this, you can tell how to answer this.1. The pattern is in the above answer. ↑** This can also include NEGATIVE EXPONENTS.I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.