Question Video: The Primitive Cubic Root of Unity | Nagwa Question Video: The Primitive Cubic Root of Unity | Nagwa

# Question Video: The Primitive Cubic Root of Unity Mathematics • Third Year of Secondary School

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Let π be the primitive cubic root of unity. 1. Find π + πΒ². 2. Find π β πΒ². 3. What is π + 1 and how is it related to the other roots of unity? 4. What is πΒ² + 1 and how is it related to the other roots of unity?

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### Video Transcript

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

To answer part one, we could try writing π and π squared in algebraic form and finding their sum that way. Alternatively, we recall that π squared is the same as the conjugate of π. And this means that π plus π squared is equal to π plus the conjugate of π. 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 plus the conjugate of π is two times the real part of π.

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 plus π squared is equal to negative one. This also means that π squared plus π plus one is equal to zero. Notice that these are the three cubic roots of unity. And weβve shown that their sum is equal to zero.

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

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

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