Question Video: Relation between Arbitrary Roots and Roots of Unity Mathematics

1) Find the solutions to the equation 𝑧⁢ = 125𝑒^((2πœ‹/3)𝑖). What are their geometrical properties? 2) State the 6th roots of unity. 3) What is the relationship between the 6th roots of unity and the solutions to the equation 𝑧⁢ = 125𝑒^((2πœ‹/3)𝑖)?

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Video Transcript

1) Find the solutions to the equation 𝑧 to the power of six equals 125𝑒 to the two πœ‹ by three 𝑖. What are their geometrical properties? 2) State the sixth roots of unity. And 3) What is the relationship between the sixth roots of unity and the solutions to the equation 𝑧 to the power of six equals 125𝑒 to the two πœ‹ by three 𝑖?

Here we have an equation involving finding the roots of a complex number. To solve this equation, we’re going to need to take the sixth root of both sides. And to do this, we’ll need to apply De Moivre’s theorem for roots. That tells us that the solutions to this equation are given by 125 to the power of one-sixth times 𝑒 to the power of two πœ‹ over three plus two πœ‹π‘˜ over six 𝑖, where π‘˜ takes values from zero through to five.

Substituting these values of π‘˜ into our formula and then subtracting multiples of two πœ‹ where necessary from the argument to express the argument within the range for the principal argument. And we see that our solutions to the equation are root five 𝑒 to the πœ‹ by nine 𝑖, root five 𝑒 to the four πœ‹ by nine 𝑖, root five 𝑒 to the seven πœ‹ by nine 𝑖, root five 𝑒 to the negative eight πœ‹ by nine 𝑖, root five 𝑒 to the negative five πœ‹ by nine 𝑖, and root five 𝑒 to the negative two πœ‹ by nine 𝑖.

As expected, when we plot these on an Argand diagram, we see that they form the vertices of a regular hexagon. And this hexagon is inscribed within a circle whose centre is the origin and whose radius is root five.

Now as we go ahead and answer part two and three of this question, we’ll leave the Argand diagram in place. It’s going to be useful to us in a moment. In a similar way, we could use De Moivre’s theorem to find the sixth roots of unity. Or we can simply recall that they are one, 𝑒 to the πœ‹ by three 𝑖, 𝑒 to the two πœ‹ by three 𝑖, negative one, 𝑒 to the negative two πœ‹ by three 𝑖, and 𝑒 to the negative πœ‹ by three 𝑖.

So to find the relationship between the sixth roots of unity and the solutions to our equation, let’s recall the geometric interpretation of the sixth roots of unity. The sixth roots of unity are represented geometrically by the vertices of a regular hexagon. This time, that hexagon is inscribed within a unit circle. And once again, its centre is the origin. And we can see that we can transform the sixth roots of unity to the roots of our equation by a dilation scale factor root five and a counterclockwise rotation by πœ‹ by nine radians.

Another way of thinking about this is it’s the same as multiplying them by a complex number whose modulus is the square root of five and whose argument is πœ‹ by nine, In other words, root five 𝑒 to the πœ‹ by nine 𝑖. This means if we call the sixth roots of unity one, πœ”, πœ” squared all the way through to πœ” to the power of five, then the roots of our equation can be expressed as root five 𝑒 to the πœ‹ by nine 𝑖, πœ” times root five 𝑒 to the πœ‹ by nine 𝑖, all the way through to πœ” to the power of five times root five 𝑒 to the πœ‹ by nine 𝑖. And these results would have stood had we used any of the other roots of 𝑧 to the power of six equals 125𝑒 to the two πœ‹ by three 𝑖.

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