Hence , and hence . Since we know is positive on , we conclude that

Observe that if
, then the problem of
calculating and is
the same as the problem of calculating . Let
. We know that the
complex solutions of are

so if we can express the solutions to in algebraic terms, then we can express and in algebraic terms. We have

Here and are obvious sixth roots of , and the other four roots are the solutions of the quadratic equations

Identify each solution with one of these exponentials. Find and .

to find and .

The numbers and can also be expressed algebraically.

If
, then , so

and since ,

The fact that says and , so

i.e.,

or

Now for all ,

so

Hence satisfies a quadratic equation.

- a) Solve (12.69), and determine and in algebraic terms.
- b) The quadratic equation has two solutions, one of which is . What is the geometrical significance of the other solution?

For example, since , we can construct a dodecagon as follows:

In the figure, make an arc of radius with center at , intersecting the -axis at . Then , so if bisects , then , and the vertical line through intersects the circle at where is a side of the -gon.

Use the formula for to inscribe a regular pentagon in a circle.

Gauss discovered this result in 1796 [31, p 754] when he was a nineteen year old student at Göttingen. The result is [21, p 458]