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Question Video: ο»ΏReciprocals of the 𝑛th Roots of Unity Mathematics

Let πœ” be an 𝑛th root of unity. Which of the following is the correct relationship between πœ” to the power negative one and πœ”? [A] πœ”β»ΒΉ = βˆ’πœ” [B] πœ”β»ΒΉ = (πœ”)^* [C] πœ”β»ΒΉ = πœ” [D] πœ”β»ΒΉ = βˆ’(πœ”)^*

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Video Transcript

Let πœ” be an 𝑛th root of unity. Which of the following is the correct relationship between πœ” to the power negative one and πœ”? Is it (A) πœ” to the power negative one is negative πœ”? (B) πœ” to the power negative one is the complex conjugate of πœ”. (C) πœ” to the power negative one is actually πœ”. Or (D) πœ” to the power negative one is the negative complex conjugate of πœ”. Express πœ” to the power negative one in terms of positive powers of πœ”.

We’re told that πœ”, which is a complex number, is an 𝑛th root of unity. And recall that if complex number πœ” is an 𝑛th root of unity, then it’s a solution to the equation πœ” to the power 𝑛 is equal to one. That’s for a positive integer 𝑛. And there are two parts to this question. The first is what’s the correct relationship between πœ” to the power negative one and πœ”, and we’re then asked to express πœ” to the power negative one in terms of positive powers of πœ”. So let’s look first at the first part.

We know that πœ” is an 𝑛th root of unity. And recall it from de Moivre’s theorem, we can express these roots in exponential form; that is, πœ” is 𝑒 to the π‘–πœƒ, where πœƒ is two π‘˜πœ‹ over 𝑛 and π‘˜ takes values from zero to 𝑛 minus one. Since we want to know the relationship between πœ” and πœ” to the power negative one, we can also express πœ” to the power negative one in exponential form. πœ” to the power negative one is equal to 𝑒 to the power negative π‘–πœƒ, where 𝑒 is Euler’s number. And that’s equal to 𝑒 to the power 𝑖 times negative πœƒ. And now, if we look at this in trigonometric form, we have πœ” to the power negative one is the cos of negative πœƒ plus 𝑖 times the sin of negative πœƒ.

But now we can use the fact that for an angle πœƒ, the cos of negative πœƒ is the cos of πœƒ and the sin of negative πœƒ is negative the sin of πœƒ so that πœ” to the power negative one is the cos of πœƒ minus 𝑖 sin πœƒ in trigonometric form. And this is simply the complex conjugate of πœ” since πœ” is cos πœƒ plus 𝑖 sin πœƒ in trigonometric form. So in fact, we have πœ” to the power negative one is the complex conjugate of πœ”; that is, option (B) πœ” to the negative one is equal to the complex conjugate of πœ”.

We can check that none of the other options apply. So, for example, option (A) says that πœ” to the negative one is equal to negative πœ”. But we can see, in fact, from the exponential form that this is not true, so we can discount option (A). Option (C) is that πœ” to the negative one is equal to πœ”. This again is not true because we have 𝑒 to the negative π‘–πœƒ against 𝑒 to the π‘–πœƒ, so these are not equal and we can discount option (C). And option (D) says that πœ” to the negative one is the negative complex conjugate of πœ”. But the negative complex conjugate of πœ” is negative cos πœƒ minus 𝑖 sin πœƒ, which is negative one times cos πœƒ plus 𝑖 sin πœƒ, which is not equal to πœ” to the power negative one. So we can discount option (D).

Now, for the second part of our question, we’re asked to express πœ” to the power negative one in terms of positive powers of πœ”. We could look again at the exponential form, but let’s instead look at what we mean by πœ” to the power negative one in terms of powers or exponents. We know that the laws of exponents for real numbers tell us that π‘Ž to the power negative one is one over π‘Ž if π‘Ž is a real number. And in fact, this applies also to complex numbers. So in fact, πœ” to the power negative one is equal to one over πœ”. And, of course, πœ” is never zero since zero is never a root of unity.

But let’s look again at our equation πœ” to the power 𝑛 is equal to one. If we divide through by πœ”, we have πœ” to the power 𝑛 over πœ” is one over πœ”. But that’s equal to πœ” to the negative one. And now if we were to write this out in full, we have πœ” to the power 𝑛 over πœ” is πœ” times itself 𝑛 times over πœ”. And we can simple this by canceling the πœ” in the denominator with one on the top. So now, in our numerator, we have πœ” times itself 𝑛 minus one times and one in the denominator, which is actually πœ” to the power 𝑛 minus one so that πœ” to the power negative one is πœ” to the power 𝑛 minus one.

Remember that 𝑛 is a positive integer so that for any 𝑛 greater than one, we have πœ” to the negative one in terms of a positive power of πœ”. And if 𝑛 is equal to one, we have 𝑛 minus one equal to zero. And in this case, in our equation, πœ” to the power 𝑛 is equal to one; that’s the 𝑛th root of unity. That means πœ” is the first root of unity, which is one. And so if πœ” is an 𝑛th root of unity, πœ” to the power negative one is the complex conjugate of πœ”, that’s option (B), and πœ” to the power negative one in terms of positive powers of πœ” is πœ” to the power 𝑛 minus one.

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