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.