# Question Video: Finding the Sum of the πth Roots of Unity Mathematics

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, theyβre the values of π§ such that π§ to the sixth power equals one. And we use a formula for finding the πth roots of unity. The polar coordinates will be cos of two ππ over π plus π sin of two ππ over π, where π is the integer values between zero and π minus one.

As weβre dealing with the sixth roots of unity, our π-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 π sin of zero. And we know that this equals one. For π equals one, we have cos of two π over six plus π sin of two π over six. We can simplify that argument to π over three. And at this point, letβs just leave this root in polar form and find the root for π equals two, which is cos of four π over six plus π sin of four π over six. That argument simplifies to two π over three. Again, weβll leave that there.

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

We can find an equivalent value inside the range by subtracting two π. When we do that, we get negative two-thirds π so that we have cos of negative two-thirds π plus π sin of negative two-thirds π as our fifth root. And weβll move on to our final sixth root, where π equals five, which is cos of 10π over six plus π sin of 10π over six. We can simplify the argument to five π over three. However, itβs outside the range of the principal argument, which means again weβll subtract two π from five π over three to get a new value of negative π over three, and our final root as cos of negative π over three plus π sin of negative π over three.

Here we have the sixth roots of unity listed in polar form. To sum these values, letβ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 π. The third root is negative one-half plus the square root of three over two π. The fourth root is the same, negative one. The fifth root is negative one-half minus the square root of three over two π, and the final root one-half minus the square root of three over two π.

Our final step will be summing these values. When we add all sixth of these terms, we see that weβ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 π minus the square root of three over two π. When we combine those values, we get zero. And again, our final term is positive square root of three over two π minus the square root of three over two π, which equals zero. The sum of the sixth roots of unity equals zero.

However, the process of calculating the πth roots of unity and then summing them is not something that you would do over and over again. And thatβs because we have a principle that tells us the sum of the πth roots of unity for π 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 weβve proven this is true by our calculations.