# Primitive roots of unity

## Homework Statement

I must show that cos(2pi/n) + isin(2pi/n) is a primitive root of unity

## Homework Equations

a primitive root of unity is an nth root of unity that does not equal 1 when raised to the kth power for k less than n and great than or equal to 1

## The Attempt at a Solution

If we set z = cos(2pi/n) + isin(2pi/n) then z^k cannot equal 1. we can use de moivres theorem to make z^k = cos(2kpi/n) + isin(2kpi/n) and then i'm not certain what fact to use next

lanedance
Homework Helper
$$cos(\frac{2 \pi}{n}} ) + i sin(\frac{2 \pi}{n}} ) = e^{\frac{2 \pi}{n}}$$

so then e^(2kpi/n) would be what i get. From there all i would need to show is that 2kpi/n cannot be zero, correct?

Char. Limit
Gold Member
$$cos(\frac{2 \pi}{n}} ) + i sin(\frac{2 \pi}{n}} ) = e^{\frac{2 \pi}{n}}$$

But that's not true. You're off by a factor of i in the exponential.

right, but to prove that z^k is not equal to zero i would have e^(i2πk/n) and this cannot equal 1 if we restrict k to be greater than or equal to 1 and less than n. So n cannot be zero and k cannot be zero, so i2(pi)k/n cannot be zero, thus z^k cannot equal 1 and is therefore primitive. and k is not equal to n so we can never have e^(i2(pi)). but how would i prove that the exponent can never be pi

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lanedance
Homework Helper
yeah cheers mis-typed the i

well its a given that k<n, so k/n < 1, so (k/n)2.pi <2.pi, isn't that enough

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no because sin(pi) is also zero and cos(pi) is 1 giving us 1 which is exactly what I have to prove cannot happen. For example choose k to be 2 and n to be 4, then we have e^(i(pi)) which is 1. I am not sure how to prove that this cannot happen though. I need to show that it is impossible to have any multiple of pi as my argument. any ideas?

Char. Limit
Gold Member
no because sin(pi) is also zero and cos(pi) is 1 giving us 1 which is exactly what I have to prove cannot happen. For example choose k to be 2 and n to be 4, then we have e^(i(pi)) which is 1. I am not sure how to prove that this cannot happen though. I need to show that it is impossible to have any multiple of pi as my argument. any ideas?

You're doing something wrong if you think that cos(pi)=1. cos(pi)=-1.

I wasn't thinking of course, you're right. I have it all proved now