Simplify each power of i. 1/i¹⁷

Question

Accepted Answer

Hello everyone. We're going to simplify the expression one over I. To the 53rd, the one over I to the 53rd to make this a little simpler, I'm going to recall that I squared is the same thing as negative one. So that means I can break down I to the 53rd into something that has an even exponent. Like I to the 52nd times I. And then I'm going to break it down one step further and make eye to the 50 seconds into eye to the second. Power to another exponent. When we raise the power to a power we multiply our exponents. So here we're gonna do the opposite and divide so I to the 52nd is the same as I squared to the 26th times I I'm gonna substitute negative one in for I squared I get one over negative one to the 26th times I because it's an even exponent. I know that this is going to come out to a positive one positive one times I it's just I. So right now I have one over. I we can't leave it there because we don't want to leave an imaginary number in our denominator. We get rid of imaginary numbers in the denominator by multiplying top and bottom by eye. So that way we have an I squared. So what this looks like one times I is I over I times I which is I squared again back to that key term that I squared equals negative one. So I now have I over negative one and recall that when you have a fraction, if there's a negative in the denominator, we can move it to the numerator. And if we have a one in the denominator, we don't have to rewrite it. So we end up with negative I as our solution, which is answer choice. D. Have a nice day.

Simplify each power of i. 1/i¹⁷

Key Concepts

Imaginary Unit i and its Powers

Negative and Fractional Exponents

Modular Arithmetic for Exponent Reduction

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Simplify each power of i. 1/i17

Key Concepts

Imaginary Unit i and its Powers

Negative and Fractional Exponents

Modular Arithmetic for Exponent Reduction

Watch next

Simplify each power of i. 1/i¹⁷