Question Video: Reciprocals of the 𝑛th Roots of Unity Mathematics

Video Transcript

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

We’re told that 𝜔, which is a complex number, is an 𝑛th root of unity. And recall that if complex number 𝜔 is an 𝑛th root of unity, then it’s a solution to the equation 𝜔 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 𝜔 to the power negative one and 𝜔, and we’re then asked to express 𝜔 to the power negative one in terms of positive powers of 𝜔. So let’s look first at the first part.

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

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

We can check that none of the other options apply. So, for example, option (A) says that 𝜔 to the negative one is equal to negative 𝜔. 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 𝜔 to the negative one is equal to 𝜔. This again is not true because we have 𝑒 to the negative 𝑖𝜃 against 𝑒 to the 𝑖𝜃, so these are not equal and we can discount option (C). And option (D) says that 𝜔 to the negative one is the negative complex conjugate of 𝜔. But the negative complex conjugate of 𝜔 is negative cos 𝜃 minus 𝑖 sin 𝜃, which is negative one times cos 𝜃 plus 𝑖 sin 𝜃, which is not equal to 𝜔 to the power negative one. So we can discount option (D).

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

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

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