Thread: Roots of Unity represented on a regular polygon - lenght of side.

Hello,

I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.

I did all of that , but I have a critical question ,

is there a relation ship between the power n and the length of the polygon side?

I mean , for the euqation Z^3 - 1 = 0 for example, is there a formula relating the power to the lenght of one side of the polygon ?
Thanks.

For each value of n, how is the length of the side of a polygon related to the central angle subtended by the side? (The side is a chord of the unit circle.)

Originally Posted by Hyunqul

I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.

There are n nth roots of unity.
.
The length of the side of the polygon is .

Thank you very much for your response,

But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?

Originally Posted by Hyunqul

Thank you very much for your response,
But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?

O.K. Let then

Thanks alot sir , I highly appreciate you help.

If I could ask one more question ... is there a proof for this rule ? The one relating roots of unity to the length of one side.

1 can be represented in polar form in the obvious way: . The nth roots are given by for k= 0 to n- 1. That is, 1 has n nth roots, equally spaced around the unit circle. If you draw lines from each root to the origin, you have n isosceles triangles and the angle at the vertex (the origin) is . If you draw a perpendicular from the origin to one side, it divides that isosceles triangle into two right triangles having hypotenuse 1 and angle . The "opposite side" to that angle is given by [itex]sin(\frac{\pi}{n})[/tex] and is 1/2 the length of one side of the polygon. The length of one side, then, is .

As a check note that if n= 4, the polygon is a square, having diagonals of length 2. while, by the Pythagorean theorem so , .

Similarly, if n= 6, the polygon is a hexagon having diagonals of length 2. .

Thanks for your generous contribution sir ,

In the same context , I was asked to obtain solutions for the equation Z^n-i = 0 for three cases where n = 1,2,3

I did that for all of them , for example when n = 3 the solutions set was :
i^(1/3) = 1^(1/3) e^[i(π/2 + 2kπ)]/3 = 1 e^[i(π/6 + 2kπ/3)] , where k = 0, 1, and 2 ... and I followed the same approach for n = 4 and 5.
Then I had to plot each of those roots on an argand diagram , and I did that as well.

After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks

Originally Posted by Hyunqul

After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks

First let .

Then where

Recall that

Yes , sir..thanks but I would like to calrify the question more :

First , it is required to generalize and prove the resuls for z when z^n = cos (x) + isin (X)

Secondly . it is required to generalize for Z when z^n = cos (Kx) + isin (Kx)

Thanks and sorry for disturbance.

Sorry I made a mistake ...

First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)

Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 )

Sorry again.

Originally Posted by Hyunqul

First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)
Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 )

Why don't you just read about this topic?