# Primitive roots of unity

• Driessen12
In summary, the homework statement is trying to find a primitive root of unity and it states that if we set z = cos(2pi/n) + isin(2pi/n) then z^k cannot equal 1. If we use de moivres theorem to make z^k = cos(2kpi/n) + isin(2kpi/n), then I'm not certain what fact to use to write it as. However, if we set k<n, then k/n < 1, so (k/n)2.pi <2.pi, isn't that enough? No because sin(pi) is also zero and cos(pi) is 1 giving us 1 whichf

## 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

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

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?

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

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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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?

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