# Question Video: Finding the 𝑛th Roots of Unity Mathematics

Find the ninth roots of unity.

03:34

### Video Transcript

Find the ninth roots of unity.

To find the ninth roots of unity, we’re trying to solve all the values of 𝑧 such that 𝑧 to the ninth power equals one. And we know a formula for calculating 𝑛th roots of unity will be cos of two 𝜋𝑘 over 𝑛 plus 𝑖 times sin of two 𝜋𝑘 over 𝑛 which is equal to 𝑒 to the two 𝜋𝑘 over 𝑛 times 𝑖 power, for 𝑘-values beginning at zero and ending with 𝑛 minus one. Since we’re dealing with the ninth root, our 𝑛 is equal to nine. And to find the ninth roots of unity, we’ll start with 𝑘 equals zero. Our first term becomes cos of zero plus 𝑖 sin of zero, which would be 𝑒 to the zero power, which is equal to one.

Next, we have 𝑘 equals one. If we plug in one for 𝑘 and nine for 𝑛, we can simplify this argument to two-ninths 𝜋, which gives us 𝑒 to the two-ninths 𝜋𝑖 power. Now for 𝑘 equals two, we get four-ninths 𝜋, which will give us 𝑒 to the four-ninths 𝜋𝑖 power. For 𝑘 equals three, our argument is six-ninths 𝜋, which we can reduce to two-thirds 𝜋, which will give us 𝑒 to the two-thirds 𝜋𝑖 power. Next, we have 𝑘 equals four. When we plug in four for 𝑘 and nine for 𝑛, we get eight-ninths 𝜋, which gives us 𝑒 to the eight-ninths 𝜋 times 𝑖 power.

At this point when we get to 𝑘 equals five, we get cos of ten-ninths 𝜋 plus 𝑖 sin of ten-ninths 𝜋. The problem here is that ten-ninths 𝜋 is outside the range for the principal argument. We want the principal argument to be greater than negative 𝜋 and less than or equal to 𝜋. And that means we need to take ten-ninths 𝜋 and subtract two 𝜋, which we write as eighteen-ninths 𝜋. When we do this subtraction, we get negative eight-ninths 𝜋. We want to substitute that value in in place of the ten-ninths 𝜋, which will give us 𝑒 to the negative eight-ninths 𝜋𝑖.

Notice the interesting pattern here. The fourth root was 𝑒 to the eight-ninths 𝜋𝑖, while the fifth root was 𝑒 to the negative eight-ninths 𝜋𝑖. Continuing on, we have 𝑘 to the sixth. And our argument is twelve-ninths 𝜋, which simplifies to four-thirds 𝜋. But again, this is outside the range of the principal argument, which means we would need to subtract two 𝜋 from the four-thirds 𝜋. When we do that, we get negative two-thirds 𝜋 so that we have 𝑒 to the negative two-thirds 𝜋𝑖.

And again we have this pattern; 𝑘 equals three gave us 𝑒 to the positive two-thirds 𝜋𝑖, and 𝑘 equals six gave us 𝑒 to the negative two-thirds 𝜋𝑖. Using this pattern for 𝑘 equals seven, we can expect that it would be equal to 𝑒 to the negative four 𝜋 over nine 𝑖. If we plugged in 𝑘 equals seven, we get a 14 over nine 𝜋. And by subtracting two 𝜋, we get negative four-ninths 𝜋. We can use this pattern to solve for 𝑘 equals eight. When 𝑘 equals eight, we have a root of 𝑒 to the negative two-ninths 𝜋𝑖. From here we wanna list out these roots in order beginning with 𝑘 equals zero, which would give us the list that looks like this.