Video: Finding the 𝑛th Roots of Unity

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 πœ‹π‘˜ over 𝑛 plus 𝑖 times sin of two πœ‹π‘˜ over 𝑛 which is equal to 𝑒 to the two πœ‹π‘˜ over 𝑛 times 𝑖 power, for π‘˜-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 π‘˜ 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 π‘˜ equals one. If we plug in one for π‘˜ and nine for 𝑛, we can simplify this argument to two-ninths πœ‹, which gives us 𝑒 to the two-ninths πœ‹π‘– power. Now for π‘˜ equals two, we get four-ninths πœ‹, which will give us 𝑒 to the four-ninths πœ‹π‘– power. For π‘˜ equals three, our argument is six-ninths πœ‹, which we can reduce to two-thirds πœ‹, which will give us 𝑒 to the two-thirds πœ‹π‘– power. Next, we have π‘˜ equals four. When we plug in four for π‘˜ and nine for 𝑛, we get eight-ninths πœ‹, which gives us 𝑒 to the eight-ninths πœ‹ times 𝑖 power.

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

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

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

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