Question Video: Finding the 𝑛th Roots of Unity | Nagwa Question Video: Finding the 𝑛th Roots of Unity | Nagwa

# Question Video: Finding the πth Roots of Unity Mathematics • Third Year of Secondary School

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Find the ninth roots of unity.

03:34

### Video Transcript

Find the ninth roots of unity.

To find the ninth roots of unity, weβre trying to solve all the values of π§ such that π§ to the ninth power equals one. And we know a formula for calculating πth roots of unity will be cos of two ππ over π plus π times sin of two ππ over π which is equal to π to the two ππ over π times π power, for π-values beginning at zero and ending with π minus one. Since weβre dealing with the ninth root, our π is equal to nine. And to find the ninth roots of unity, weβll start with π equals zero. Our first term becomes cos of zero plus π sin of zero, which would be π 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 π, we can simplify this argument to two-ninths π, which gives us π to the two-ninths ππ power. Now for π equals two, we get four-ninths π, which will give us π to the four-ninths ππ power. For π equals three, our argument is six-ninths π, which we can reduce to two-thirds π, which will give us π to the two-thirds ππ power. Next, we have π equals four. When we plug in four for π and nine for π, we get eight-ninths π, which gives us π to the eight-ninths π times π power.

At this point when we get to π equals five, we get cos of ten-ninths π plus π 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 π to the negative eight-ninths ππ.

Notice the interesting pattern here. The fourth root was π to the eight-ninths ππ, while the fifth root was π to the negative eight-ninths ππ. Continuing on, we have π 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 π to the negative two-thirds ππ.

And again we have this pattern; π equals three gave us π to the positive two-thirds ππ, and π equals six gave us π to the negative two-thirds ππ. Using this pattern for π equals seven, we can expect that it would be equal to π to the negative four π over nine π. If we plugged in π 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 π equals eight. When π equals eight, we have a root of π to the negative two-ninths π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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