Video: Finding the Sum of the 𝑛th Roots of Unity

Find the sum of the sixth roots of unity.

03:58

Video Transcript

Find the sum of the sixth roots of unity.

When we think about the sixth roots of unity, they’re the values of 𝑧 such that 𝑧 to the sixth power equals one. And we use a formula for finding the 𝑛th roots of unity. The polar coordinates will be cos of two πœ‹π‘˜ over 𝑛 plus 𝑖 sin of two πœ‹π‘˜ over 𝑛, where π‘˜ is the integer values between zero and 𝑛 minus one.

As we’re dealing with the sixth roots of unity, our 𝑛-value equals six. And our sixth roots will come from π‘˜-values between zero and five. When π‘˜ equals zero, we have cos of zero plus 𝑖 sin of zero. And we know that this equals one. For π‘˜ equals one, we have cos of two πœ‹ over six plus 𝑖 sin of two πœ‹ over six. We can simplify that argument to πœ‹ over three. And at this point, let’s just leave this root in polar form and find the root for π‘˜ equals two, which is cos of four πœ‹ over six plus 𝑖 sin of four πœ‹ over six. That argument simplifies to two πœ‹ over three. Again, we’ll leave that there.

Then we have the cos of three πœ‹ over three plus 𝑖 sin of three πœ‹ over three. The cos of πœ‹ plus the 𝑖 sin of πœ‹ equals negative one. For our fifth root, we have the cos of eight πœ‹ over six plus the 𝑖 sin of eight πœ‹ over six. We know that this argument simplifies to four πœ‹ over three but that four πœ‹ over three is outside the range of the principal argument.

We can find an equivalent value inside the range by subtracting two πœ‹. When we do that, we get negative two-thirds πœ‹ so that we have cos of negative two-thirds πœ‹ plus 𝑖 sin of negative two-thirds πœ‹ as our fifth root. And we’ll move on to our final sixth root, where π‘˜ equals five, which is cos of 10πœ‹ over six plus 𝑖 sin of 10πœ‹ over six. We can simplify the argument to five πœ‹ over three. However, it’s outside the range of the principal argument, which means again we’ll subtract two πœ‹ from five πœ‹ over three to get a new value of negative πœ‹ over three, and our final root as cos of negative πœ‹ over three plus 𝑖 sin of negative πœ‹ over three.

Here we have the sixth roots of unity listed in polar form. To sum these values, let’s rewrite them in algebraic form. The one stays the same. The second root becomes one-half plus the square root of three over two 𝑖. The third root is negative one-half plus the square root of three over two 𝑖. The fourth root is the same, negative one. The fifth root is negative one-half minus the square root of three over two 𝑖, and the final root one-half minus the square root of three over two 𝑖.

Our final step will be summing these values. When we add all sixth of these terms, we see that we’re adding one plus negative one, which equals zero. One-half plus negative one-half equals zero. Again, negative one-half plus one-half equals zero. From there, we have the square root of three over two 𝑖 minus the square root of three over two 𝑖. When we combine those values, we get zero. And again, our final term is positive square root of three over two 𝑖 minus the square root of three over two 𝑖, which equals zero. The sum of the sixth roots of unity equals zero.

However, the process of calculating the 𝑛th roots of unity and then summing them is not something that you would do over and over again. And that’s because we have a principle that tells us the sum of the 𝑛th roots of unity for 𝑛 is greater than one is zero. By this principle, we can say that the sum of the sixth roots of unity will be equal to zero. And we’ve proven this is true by our calculations.

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