Using De Moivre Formula to Prove z^n = 1 is a Root of Unity

[Bob19]

Hi
I have a question regarding the use of de moivre formula:
I'm presented with a complex number $$z = (cos(v) + i sin(v))^n = 1$$
I'm suppose to show that $$z^n = 1$$ is a root of unity. Is there a procedure on how to show this? If n = 6 and $$v = \frac{4 \pi}{6}$$

Sincerely and Best Regards

Bob

 

[HallsofIvy]

Are you sure you've copied this correctly? You are saying that you are told that z= 1 ?? And you want to prove that 1 is a "root of unity"??

I very much doubt that! Please copy the problem carefully.

Perhaps you are given that z= cos v+ i sin(v) and want to show that (cos v+ i sin(v))ⁿ= zⁿ= 1, thus showing that z is a "root of unity".

Furthermore, you then say "if n= 6 and $v= \frac{4\pi}{6}$". Is that a separate problem or is the original problem to show that
$$\left(cos\left(\frac{4\pi}{6})+ i sin(\frac{4\pi}{6}\right)\right)^6= 1$$?

If it is the latter, since YOU titled this "DeMoivre's Formula Question", what does DeMoivre's formula tell you $\left(\(cos\left(\frac{4\pi}{6})+ i sin(\frac{4\pi}{6}\right)\right)^6$ is?

 

[Bob19]

Hello I have been looking through my textbook which gives the following procedure on how to show if $$(cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))$$

Is the six root of unity for $$z^{6} = 1$$

Its know that $$1 = cos(2 k \pi) + i sin(2 k \pi)$$ where k = 1,2,3,...m

According to De Moivre's formula the n'th root unity can be expressed as

$$(cos(\frac{2 k \pi}{n}) + i sin(\frac{2 k \pi}{n}))$$

My case

k = 2 and n = 6

If I insert these into De Moivre's formula I get $$z = (cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))$$

I insert z into the initial equation and get

$$(cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))^6 = (cos(\frac{24 \pi}{6}) + i sin(\frac{24 \pi}{6})) = 1$$

Therefore $$(cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))$$ is the 6'th root of unity for

$$z^6 = 1$$

I have hand it in tomorrow so therefore am I on the right track?

Sincerely and Best Regards,

Bob

 

[HallsofIvy]

Yes, you can do that but it would be easier, since you are already given $$z = (cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))$$,
to just take the 6 th power of that and show that it is 1.