In Exercises 95–99, perform the indicated operations and write th...

Question

In Exercises 95–99, perform the indicated operations and write the result in standard form.

(i^85 - i^83)/i^45

Accepted Answer

Hello. Today we're going to be rewriting a complex expression in standard form. So the expression given to us is I race to the power of 105 -1 raise to the power of 103. And that quantity is being divided by I raised to the power of 85. Okay, so in order to simplify this, notice how in our numerator we have I to the 105th power and I to the 103rd power. So both of these terms have at least I to the 103rd. So the numerator has the greatest common factor of I to the 103rd. And we can go ahead and factor that out from the numerator. Doing so is going to give us eye to the 103rd power times I squared minus one. And that quantity is still being divided by eye to the 85th power. Now we have terms of I in both the numerator and denominator so we can perform exponential division in order to simplify this. Well I to the 103rd divided by I to the 85th is the same thing as saying I to 103rd -85. And that's going to give you eye to the 18th power. So simplifying this is going to leave you with eye to the 18th of the numerator and one of the denominator. So we've reduced this expression to I to the 18th times I squared minus one. Now keep in mind ideally we want to work with ice squared because I square can be rewritten as negative one. While I to the 18th can also be rewritten as I squared, raised to the power of nine. So if we rewrite this, what we have is I squared race to the ninth. Power times the quantity I squared minus one. And since I squared minus one is equal to negative one, we can go ahead and replace the ice squares with negative one. Doing so is going to give us negative one, raised to the ninth. Power times the quantity negative one minus one. Now negative one. Race to the ninth. Power is just negative one. So what we have is negative one times negative one minus one, which is going to be negative two and negative one times negative two is going to give us positive two. And that is going to be the standard form of our complex expression. And the answer to this problem is going to be C. So I hope this video helps you in understanding how to write the complex expression in standard form. And I'll go ahead and see you all in the next video

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