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why is the sum of all the roots of unity equal to zero?

z^(1/n)=z1,z2,...zn

z1+z2+...+zn=0

It's obviously true when there's an even number of roots, (because each root has a partner that is pi radians away and therefore the negative of the other root). but i cant figure out why this is true when there's an odd number of roots. Even stranger, the sum of all the binomial combinations of products of roots always equals zero, regardless of the number of roots that are chosen.

for example:

z^(1/4)=w1,w2,w3,w4

w1*w2+w1*w3+w1*w4+w2*w3+w2*w4+w3*w4=0

w1*w2*w3+w1*w2*w4+w1*w3*w4+w2*w3*w4=0

z^(1/5)=w1,w2,w3,w4,w5

w1*w2+w1*w3+w1*w4+w1*w5+w2*w3+w2*w4+w2*w5+w3*w4+w3*w5+w4*w5=0

w1*w2*w3+w1*w2*w4+w1*w2*w5+w1*w3*w4+w1*w3*w5+w1*w4*w5+w2*w3*w4+w2*w3*w5+w2*w4*w5+w3*w4*w5=0

w1*w2*w3*w4+w1*w2*w3*w5+w1*w2*w4*w5+w1*w3*w4*w5+w2*w3*w4*w5=0

And, for z^(1/n)=w1,w2,w3, ... ,wn:

the product of all the roots equals z if n is odd and -z if n is even.

I understand the case for positive real z, z^(1/2)=w1,w2 then w1 is negative and w2 is positive so w1*w2=-z.

It makes sense that the product of the roots should equal either positive or negative z. But what does the oddness or evenness of n have to do with the product of the roots?

Why are these rules for sums/products/sums of products true?

Also, im sure that viete's formulas must derived from these rules, but how?

Another question:

Something i've been thinking about:

can e^(ix) have more than one value?

my reasoning comes from the rule (a^b)^c=a^(b*c) and the existence of roots of unity

applying this rule to e^(ix),

e^(ix)=(e^i)^x

e^i=cos(1)+isin(1)

given that trig functions have a period of 2pi,

(e^i)=cos(1+2pi*n)+isin(1+2pi*n) n is an integer.

therefore,

(e^i)^x=(cos(1+2pi*n)+isin(1+2pi*n))^x

(e^i)^x=cos(x+x*2pi*n)+isin(x+x*2pi*n)

e^(ix)=cos(x+x*2pi*n)+isin(x+x*2pi*n)

if x is not an integer, then e^(ix) has more than one value. if x is an irrational number, then e^(ix) is every point on the unit circle

An example:

using the approximation pi=22/7

e^(ipi)=(e^i)^(22/7)

e^(ipi)=cos(22/7+44pi/7)+isin(22/7+44pi/7)

e^(ipi)=cos(22/7+88pi/7)+isin(22/7+88pi/7)

e^(ipi)=cos(22/7+132pi/7)+isin(22/7+132pi/7)

e^(ipi)=cos(22/7+176pi/7)isin(22/7+176pi/7)

e^(ipi)=cos(22/7+220pi/7)+isin(22/7+220pi/7)

e^(ipi)=cos(22/7+264pi/7)+isin(22/7+264pi/7)

e^(ipi)=cos(22/7+44pi)+isin(22/7+44pi)

e^(ipi) has 7 values using the approximation with 7 as the bottom number, and only the last is (approximately) -1.

using the approximation pi=110000000000000/35014087480217, there are 35014087480217 values of e^(ipi), and each one is part of the unit circle.

as the fraction approaches the limit there will be infinitely many values of e^(ipi), and thus the identity e^(ipi)=-1 isnt really special.

Somebody please explain, how am i wrong? why does the rule (a^b)^c=a^(b*c) not apply to euler's formula?

Also, something that i think is related (the problem comes also comes from the same rule) but was never explained to me

Suppose i have the following equality:

x^1=x

doing the substitution

1=n*(1/n)

x^(n*(1/n))=x

using the rule (a^b)^c=a^(b*c), i get 2 equations that should be equivalent

x^1=(x^n)^(1/n)

x^1=(x^(1/n))^n

if i do the power first and the radical second, i get x and the roots of unity.

if i do the radical first and the power second, i get the roots of unity raised to the power, therefore just x.

So then (a^b)^c does not equal (a^c)^b

But by the communativity of multiplication a^(b*c)=a^(c*b), so the order of powers shouldnt matter.

Why?

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# Is e^ix multivalued, roots of unity, etc

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