Video: Equivalent Expressions for the Quintic Roots of Unity

Let πœ” be one of the quintic roots of unity. Which of the following is an equivalent expression to πœ”β»Β³? [A] πœ”Β² [B] 1/(πœ”)Β² [C] βˆ’πœ”Β³ [D] πœ”βΈ [E] πœ”β»ΒΉβ΅

08:25

Video Transcript

Let πœ” be one of the quintic roots of unity. Which of the following is an equivalent expression to πœ” to the power of negative three? Option (A) πœ” squared, option (B) one over πœ” squared, option (C) negative one times πœ” cubed, option (D) πœ” to the eighth power, or option (E) πœ” to the power of negative 15.

In this question, we’re given a number πœ” and we’re told that πœ” is one of the quintic roots of unity. We’re not told which quintic root of unity πœ” is. We need to use this information to determine an equivalent expression to πœ” to the power of negative three. And we’re given five possible options. There’s several different methods we could use to answer this question. Let’s start by recalling what we mean by a quintic root of unity. First, we recall if 𝑧 to the 𝑛th power is equal to one, then we say that 𝑧 is an 𝑛th root of unity. And we do know how to generate exact forms for all of the 𝑛th roots of unity for any positive integer 𝑛. However, it’s not actually necessary to use this to answer this question. So we’re going to do this without using this.

The question tells that πœ” is a quintic root of unity. Because this is a quintic root of unity, this tells us that our value of 𝑛 is going to be equal to five. Therefore, because πœ” is a quintic root of unity, πœ” to the fifth power has to be equal to one. We want to use this equation to determine an equivalent expression to πœ” to the power of negative three. To do this, let’s start by recalling what we mean by πœ” to the power of negative three. We’ve seen our laws of exponents for real numbers; however, a lot of these also extend to complex numbers. πœ” to the power of negative three will just be one divided by πœ” cubed. So, we want to use the fact that πœ” to the fifth power is equal to one to find an expression for one divided by πœ” cubed.

One way of doing this is just to divide both sides of our equation through by πœ” cubed. And there is one thing worth pointing out here. We do know that πœ” cannot be equal to zero because, of course, zero is never a root of unity. Zero to the power of 𝑛 for any value of 𝑛 is always equal to zero. It’s never equal to one. So by dividing through by πœ” cubed, we get πœ” to the fifth power divided by πœ” cubed is equal to one over πœ” cubed. If we wanted to use our laws of exponents, we could simplify the left-hand side. We get πœ” to the fifth power divided by πœ” cubed is equal to πœ” squared because we know when we divide these, all we would need to do is subtract the exponents.

And this is a completely valid way of simplifying this expression. However, we can actually do this directly. Instead of using our laws of exponents, we might want to write this out in full. πœ” to the fifth power will be equal to a product of πœ”β€™s, where πœ” appears five times in our product. Similarly, πœ” cubed is going to be equal to πœ” multiplied by πœ” multiplied by πœ”. And the right-hand side of this equation can be simplified by using our laws of exponents. One over π‘Ž to the 𝑛th power is just equal to π‘Ž to the power of negative 𝑛. So one over πœ” cubed is just πœ” to the power of negative three.

Well now we can directly simplify the left-hand side of our equation by canceling the shared factors of πœ” in our numerator and our denominator, and this leaves us with πœ” times πœ” divided by one. And, of course, πœ” times πœ” is equal to πœ” squared, giving us that πœ” squared is equal to πœ” to the power of negative three. And we can see that this is given as an answer in option (A).

For due diligence, we could also check all four of the other options to see if these could be correct answers as well because we don’t know there could be multiple equivalent expressions to πœ” to the power of negative three. However, we’ll see that all four of these expressions are not necessarily equivalent to πœ” to the power of negative three. And there’s a lot of different ways of doing this for each of these. First, recall that πœ” is allowed to be any of our quintic roots of unity. For the expressions to be equivalent, it must be true for all of the quintic roots of unity. And in particular, one of our quintic roots of unity would be when πœ” is equal to one. Therefore, for these expressions to be equivalent, they have to be true when πœ” is equal to one.

So let’s see what happens when πœ” is equal to one in πœ” to the power of negative three. When πœ” is equal to one, we end up with one to the power of negative three, which is of course just equal to one. And if we were to substitute πœ” is equal to one into each of our four remaining options, we would see in option (C) we end up with negative one times one cubed, which is, of course, equal to negative one. Therefore, option (C) cannot be correct. It’s not equivalent to πœ” to the power of negative three when πœ” is equal to one. To finish answering this question, we’re going to want to simplify the remaining three options. And to do this, we’re going to need to use our definition of πœ”. πœ” to the fifth power is equal to one.

Let’s start with option (D) πœ” to the eighth power. Of course, we know raising a number to the eighth power means we multiply it by itself where our product contains eight πœ”β€™s. And we, of course, know this is the same as πœ” to the fifth power multiplied by πœ” cubed. And we could’ve skipped directly to this step by using our laws of exponents. But remember, πœ” is a fifth root of unity. πœ” to the fifth power will just be equal to one. So in fact, option (D) πœ” to the eighth power is just equal to πœ” cubed.

We can do something similar for options (B) and (E). In option (E), πœ” to the power of negative 15 is one over πœ” to the 15th power. We could write this out in full and simplify. Or by using our laws of exponents, we can just show that πœ” to the 15th power is πœ” to the fifth power all cubed. And once again, πœ” is a quintic root of unity. So πœ” to the fifth power is just equal to one. So this simplifies to just give us one over one cubed, which is of course just equal to one. So, in fact, the expression in option (E) is just equal to one.

Finally, we’ll do something similar for our last option. One over πœ” squared will be equal to πœ” to the fifth power divided by πœ” squared because, remember, πœ” to the fifth power is just equal to one. And then either by writing this out in full and canceling or by using our laws of exponents, we can see this is equal to πœ” cubed. We just subtract the exponents of πœ”. So now we’ve shown something interesting. Both option (D) and option (B) are equal to πœ” cubed and option (E) is just equal to one.

Now there’s a few different ways of showing all three of these options are not correct. We’re just going to do this by finding a fifth root of unity which shows our expressions are not equivalent. And to do this, we’re going to use a simplified version of our rule for generating roots of unity. We know 𝑒 to the power of 𝑖 times two πœ‹ by 𝑛 will be an 𝑛th root of unity. And we do know this is true if we multiply our exponent by π‘˜. This will be a root of unity for any integer value of π‘˜. But we don’t need to use this; in fact, we’re just going to set our value of π‘˜ equal to one. We’re looking for quintic roots of unity, so we’ll set our value of 𝑛 equal to five.

So let’s let 𝑧 be the complex number 𝑒 to the power of two πœ‹ by five multiplied by 𝑖. So 𝑧 is one of our fifth roots of unity. So for our expressions to be equivalent, they have to be true when πœ” is equal to 𝑧. To do this, let’s start by calculating expressions for 𝑧 cubed and 𝑧 to the power of negative three. 𝑧 cubed is just going to be equal to 𝑒 to the power of two πœ‹ by five times 𝑖 all cubed. And then we can simplify this either by writing this out in full or by using our laws of exponents. In both scenarios, we end up multiplying the exponent of 𝑒 by three. This gives us 𝑒 to the power of six πœ‹ by five times 𝑖.

We can then do something very similar to find 𝑧 to the power of negative three. It’s equal to 𝑒 to the power of negative six πœ‹ by five times 𝑖. And now we’re finally ready to show that all three of our options are not correct. We want to write out these expressions in terms of their polar form. And, remember, to do this, the argument of the cosine and sine function will be our coefficient of 𝑖 in our exponent. In other words, 𝑧 cubed is the cos of six πœ‹ by five plus 𝑖 times the sin of six πœ‹ by five, and 𝑧 to the power of negative three is the cos of negative six πœ‹ by five plus 𝑖 times the sin of negative six πœ‹ by five.

And for these two expressions to be equal, their real and imaginary parts must be equal. If we were to type these into a calculator, we would see that their imaginary parts are not equal. The sin of six πœ‹ by five is not equal to the sin of negative six πœ‹ by five. But remember, 𝑧 is one of our quintic roots of unity. So what we’ve shown here is πœ” cubed is not always equal to πœ” to the power of negative three. And using a very similar set of reasoning, we can see 𝑧 to the power of negative three is not equal to one. For example, its real component is not equal to one. So option (E) cannot be correct either because 𝑧 is a quintic root of unity and 𝑧 to the power of negative three is not equal to one.

Therefore, we were able to show if πœ” is a quintic root of unity, then, of the five given expressions, the only one equivalent to πœ” to the power of negative three is πœ” squared which was option (A).

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