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

1) Find the solutions to the equation π§βΆ = 125π^((2π/3)π). 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 π§βΆ = 125π^((2π/3)π)?

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### Video Transcript

1) Find the solutions to the equation π§ to the power of six equals 125π to the two π by three π. 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 π§ to the power of six equals 125π to the two π by three π?

Here we have an equation involving finding the roots of a complex number. To solve this equation, weβre going to need to take the sixth root of both sides. And to do this, weβll need to apply De Moivreβs theorem for roots. That tells us that the solutions to this equation are given by 125 to the power of one-sixth times π to the power of two π over three plus two ππ over six π, where π takes values from zero through to five.

Substituting these values of π into our formula and then subtracting multiples of two π 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 π to the π by nine π, root five π to the four π by nine π, root five π to the seven π by nine π, root five π to the negative eight π by nine π, root five π to the negative five π by nine π, and root five π to the negative two π by nine π.

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, weβll leave the Argand diagram in place. Itβs going to be useful to us in a moment. In a similar way, we could use De Moivreβs theorem to find the sixth roots of unity. Or we can simply recall that they are one, π to the π by three π, π to the two π by three π, negative one, π to the negative two π by three π, and π to the negative π by three π.

So to find the relationship between the sixth roots of unity and the solutions to our equation, letβ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 by nine radians.

Another way of thinking about this is itβs the same as multiplying them by a complex number whose modulus is the square root of five and whose argument is π by nine, In other words, root five π to the π by nine π. This means if we call the sixth roots of unity one, π, π squared all the way through to π to the power of five, then the roots of our equation can be expressed as root five π to the π by nine π, π times root five π to the π by nine π, all the way through to π to the power of five times root five π to the π by nine π. And these results would have stood had we used any of the other roots of π§ to the power of six equals 125π to the two π by three π.