In Exercises 53–60, write each power of i as as i, - 1, - i, or 1...

Question

In Exercises 53–60, write each power of i as as i, - 1, - i, or 1.

i^31

Accepted Answer

Hello. Everyone in this video we're going to be looking at this practice problem where we're trying to write the expression I to the 67th power as negative 11 negative I or I. And in this case recall that we're working with I. Which is equal to the square root of -1. And we have I to the 67th. So let's try to see what happens when we add powers to I. So if we have I squared that is the same as I times I. Which is equal to the square root of negative one times the square root of negative one or negative one. If we had another power I to the third that's the same as I times I squared Not equal to the square root of - times negative one. Leaving us with negative square root of negative one. We had another power I to the fourth. The same as I squared times I squared which is equal to negative one times negative one. Leaving us with positive one. We had one more power I to the fifth. That's the same as I times I to the fourth which is equal to the square root of negative one. Times one which is equal to the square root of negative one or I. So in this case you can see that by the time we got to the 5th component we started to repeat I. So we're going to use this to try to simplify our expression and we're going to use this by seeing that I to the fourth is equal to one. So we're going to try to separate our expression. I. To the 67th into I. To the fourth to an ex opponent which I don't know yet. X. Times I. To another exponent that I don't know yet. So in this case If I'm trying to find all the times that I to the fourth can fit into I. To the 67th I need to divide by four. And if I do this you can see that four goes into 61 time and four goes into 27 six times. So in this case I have I to the fourth Going into 67 times and my remainder Is three. So I'm off with I to the third and now I can substitute one in for I to the fourth. If I do that I'm left with One to the 16th times I to the third. Well 1 to the 16th will just be one. So I really only have I to the third And if we go back to I to the 3rd here you can see that it is equal to Negative Square Root of -1. Well in this case the square root of -1 is equal to I. So this can also be negative. I. So once we simplify I to the 67th is really just negative I. So this becomes the final answer and you can see that this aligns with answer choice. C. So hopefully this video helped and see you next time.

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