Question Video: Finding the Fifth Roots of Unity Mathematics

Video Transcript

Find the fifth roots of unity.

Well, these are the five values for 𝑧 that satisfy the equation 𝑧 to the fifth equals one. And we can use de Moivre’s theorem to help us to solve this equation for 𝑧. Now, this states that for a complex number 𝑧, which is equal to 𝑟 times cos 𝜃 plus 𝑖 sin 𝜃, then the 𝑛th root of 𝑧 is given by 𝑧 to the power of one over 𝑛 is equal to 𝑟 to the power of one over 𝑛 times cos of 𝜃 plus two 𝜋𝑘 over 𝑛 plus 𝑖 sin 𝜃 plus two 𝜋𝑘 over 𝑛 for values of 𝑘 from zero, one, two, up to 𝑛 minus one.

So, looking back at our equation, we’ve got 𝑧 to the five is equal to one. And we can write one in polar form as one times cos of zero plus 𝑖 sin of zero. In other words, 𝑟 is equal to one and 𝜃 is equal to zero. Then, with 𝑟 equal to one, 𝑟 to the one over 𝑛 is just going to be equal to one. And with 𝜃 equal to zero, we can say that 𝑧 to the one over 𝑛 is equal to cos of two 𝜋𝑘 over 𝑛 plus 𝑖 sin two 𝜋𝑘 over 𝑛, where 𝑘 takes the integer values from zero to 𝑛 minus one. And we could write that in exponential form as 𝑒 to the power of two 𝜋𝑘 over 𝑛 𝑖.

So now we’ve got this general formula for finding the 𝑛th roots of unity. Let’s go back and answer our specific question. To find the fifth roots of unity, we’re going to set 𝑛 equal to five, and 𝑘 will take the integer values from zero up to five minus one. So that’s four. When 𝑘 equals zero, 𝑧 to the one over five is equal to cos of two 𝜋 times zero over five plus 𝑖 sin of two 𝜋 times zero over five. Well, two 𝜋 times zero is zero. So, that gives us cos of zero plus 𝑖 sin of zero or, in exponential form, 𝑒 to the power of zero 𝑖, which of course is just 𝑒 to the power of zero or one. So, our first root is one.

For our next root, let’s plug in 𝑘 equals one. That gives us cos of two 𝜋 times one over five plus 𝑖 sin of two 𝜋 times one over five. And, of course, two 𝜋 times one over five is just two 𝜋 over five. Then, in exponential form, our next root is 𝑒 to the power of two 𝜋 over five 𝑖. And so we carry on. When 𝑘 is equal to two, we get cos of two 𝜋 times two over five plus 𝑖 sin two 𝜋 times two over five, which simplifies to cos of four 𝜋 over five plus 𝑖 sin four 𝜋 over five, and, in exponential form, that’s 𝑒 to the power of four 𝜋 over five 𝑖. Next, we plug in 𝑘 equal to three and find that our next root is cos of six 𝜋 over five plus 𝑖 sin six 𝜋 over five.

But we generally try to give our answers so that this principal argument here is greater than negative 𝜋 but less than or equal to positive 𝜋. So let’s find an equivalent angle in that interval. Because of the periodic nature of this function, we can just subtract two 𝜋 and see if that gives us an answer in the correct interval. Well, six 𝜋 over five minus two 𝜋 is equivalent to six 𝜋 over five minus 10𝜋 over five, which is negative four 𝜋 over five. And that is in the required interval, so we can use that value instead. And the equivalent exponential form is 𝑒 to the negative four 𝜋 over five 𝑖.

Now, for our final root, we can plug in 𝑘 equals four. And this gives us cos of eight 𝜋 over five plus 𝑖 sin of eight 𝜋 over five. And again, we’ve got a value for our principal argument which is greater than 𝜋. So, if we subtract two 𝜋, we get negative two 𝜋 over five. Then our final root in exponential form is 𝑒 to the negative two 𝜋 over five 𝑖. So, our fifth roots of unity are one, 𝑒 to the two-fifths 𝜋𝑖, 𝑒 to the four-fifths 𝜋𝑖, 𝑒 to the negative four-fifths 𝜋𝑖, and 𝑒 to the negative two-fifths 𝜋𝑖.