Question Video: The Primitive Cubic Root of Unity Mathematics

Let πœ” be the primitive cubic root of unity. 1. Find πœ” + πœ”Β². 2. Find πœ” βˆ’ πœ”Β². 3. What is πœ” + 1 and how is it related to the other roots of unity? 4. What is πœ”Β² + 1 and how is it related to the other roots of unity?

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Video Transcript

Let πœ” be the primitive cubic root of unity. 1) Find πœ” plus πœ” squared. 2) Find πœ” minus πœ” squared. 3) What is πœ” plus one and how is it related to the other roots of unity? 4) What is πœ” squared plus one and how is it related to the other roots of unity?

To answer part one, we could try writing πœ” and πœ” squared in algebraic form and finding their sum that way. Alternatively, we recall that πœ” squared is the same as the conjugate of πœ”. And this means that πœ” plus πœ” squared is equal to πœ” plus the conjugate of πœ”. This has its own property. We know that the sum of a complex number and its conjugate is two times the real part of that complex number. So πœ” plus the conjugate of πœ” is two times the real part of πœ”.

Well, the real part of the primitive cubic root of unity is negative one-half. And two times negative a half is negative one. So we can see that πœ” plus πœ” squared is equal to negative one. This also means that πœ” squared plus πœ” plus one is equal to zero. Notice that these are the three cubic roots of unity. And we’ve shown that their sum is equal to zero.

Let’s repeat this process for part two. Once again, we express πœ” squared as the conjugate of πœ”. But this time, the difference between a complex number and its conjugate is two 𝑖 times the imaginary part of that complex number. The imaginary part of πœ” is root three over two. So the difference between πœ” and πœ” squared is 𝑖 root three or root three 𝑖. And we can use what we’ve calculated here to work out πœ” plus one for part three and πœ” squared plus one for part four. We saw that πœ” squared plus πœ” plus one is equal to zero. So let’s subtract πœ” squared from both sides of this equation. When we do, we see that πœ” plus one is equal to negative πœ” squared. Similarly, we can also deduce that πœ” squared plus one is equal to negative πœ”.

So the cubic roots of unity have three really important properties. We know that πœ” squared is equal to the conjugate of πœ”. We know that the sum of the three cubic roots is equal to zero. And we know that πœ” minus πœ” squared is equal to 𝑖 root three.

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