# Question Video: Geometric Interpretation of the Roots of Unity Mathematics • 12th Grade

1) Find all the solutions to 𝑧⁶ = 1. 2) By plotting the solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions of 𝑧⁶ = 1.

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### Video Transcript

1) Find all the solutions to 𝑧 to the power of six equals one. 2) By plotting the solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions of 𝑧 to the power of six equals one.

We could solve this equation by finding the sixth root of both sides. However, we know that there are going to be six solutions to this equation. So we need to consider an alternative method. Instead, we rearrange by subtracting one from both sides. And we see that 𝑧 to the power of six minus one equals zero. This is actually a special case of the difference of two squares, meaning we can write the expression on the left-hand side as 𝑧 cubed minus one multiplied by 𝑧 cubed plus one. And now, we have two numbers whose product is zero. This can only be the case if 𝑖, the number itself, is equal to zero.

Let’s start with saying that 𝑧 cubed minus one is equal to zero. We can observe that one of the solutions to this equation is one since one cubed minus one is indeed zero. This means that 𝑧 minus one must be a factor of 𝑧 cubed minus one. We could use polynomial long division to find the other factor. Or we could say that this means that 𝑧 cubed minus one is equal to 𝑧 minus one multiplied by some quadratic. And then, we can equate coefficients of 𝑧. Distributing the brackets, and we see that 𝑎𝑧 cubed plus 𝑏 minus 𝑎 𝑧 squared plus 𝑐 minus 𝑏 𝑧 minus 𝑐 equals 𝑧 cubed minus one.

Equating coefficients of 𝑧 cubed, we see that 𝑎 is equal to one. And that’s because the coefficient of 𝑧 cubed on the right-hand side is just one. The coefficient of 𝑧 squared on the right-hand side is zero. So we see that when we equate coefficients of 𝑧 squared, we get 𝑏 minus 𝑎 equals zero. 𝑎 is of course one. So 𝑏 minus one is zero, which means that 𝑏 must be equal to one. We’re going to skip equating coefficients of 𝑧 to the power of one and go straight to equating constants or coefficients of 𝑧 to the power of zero.

We see that negative 𝑐 equals negative one, which means that 𝑐 is equal to one. And this means that 𝑧 cubed minus one is equal to 𝑧 minus one multiplied by 𝑧 squared plus 𝑧 plus one. We then solve 𝑧 squared plus 𝑧 plus one equals zero by either using the quadratic formula or completing the square.

If we use the quadratic formula, we see that 𝑧 is equal to negative one plus or minus the square root of one squared minus four times one times one, all over two times one. That’s negative one plus or minus the square root of negative three over two. We’ll split this up and write it as negative one-half plus or minus the square root of negative three over two. And since the square root of negative one is 𝑖, our solutions for 𝑧 become negative a half plus or minus the square root of three over two 𝑖.

We’ll repeat this process for 𝑧 cubed plus one is equal to zero. This time, we can spot that one of the solutions to this equation is 𝑧 equals negative one. And that’s because negative one cubed plus one is equal to zero. This time, that means that 𝑧 plus one must be a factor of 𝑧 cubed plus one. And we can say that we can write 𝑧 cubed plus one as 𝑧 plus one multiplied by some quadratic in 𝑧.

This time, distributing the brackets, and we see that 𝑎𝑧 cubed plus 𝑎 plus 𝑏 𝑧 squared plus 𝑏 plus 𝑐 𝑧 plus 𝑐 equals 𝑧 cubed plus one. And this time, when we equate coefficients, we get that 𝑎 is equal to one. 𝑏 is equal to negative one. And 𝑐 is equal to one. So 𝑧 cubed plus one is equal to 𝑧 plus one multiplied by 𝑧 squared minus 𝑧 plus one. This time, we solve 𝑧 squared minus 𝑧 plus one equals zero, once again using the quadratic formula or possibly completing the square. And when we do, we can see that 𝑧 is equal to one-half plus or minus the square root of three over two 𝑖.

And we see that we now have the six solutions to the equation 𝑧 to the power of six equals one that we were looking for. And if we want to, we could check these solutions by substituting them back into the equation 𝑧 to the power of six equals one and checking that our answers make sense.

For part 2), we’re going to plot these points on an Argand diagram. 𝑧 equals one and 𝑧 equals negative one are fairly straightforward. We have the point a half, root three over two representing the solution one-half plus root three over two 𝑖. And we have negative a half root three over two, representing the solution negative a half plus root three over two 𝑖. We can plot the other two solutions as shown. And what about the geometric properties? Well, we can see that these complex numbers are evenly spaced about the origin. In fact, the solutions are the vertices of a regular hexagon inscribed with a unit circle whose centre is the origin.