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Is e^ix multivalued, roots of unity, etc

  1. Jan 9, 2009 #1

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



    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),

    given that trig functions have a period of 2pi,
    (e^i)=cos(1+2pi*n)+isin(1+2pi*n) n is an integer.

    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) 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:

    doing the substitution


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


    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.

  2. jcsd
  3. Jan 9, 2009 #2


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    You can write down a closed form formula for the sum. Hint:
    It's a geometric series

    I'm confused -- all of the previous questions are direct consequences by Viète's formulas. Are you asking how Viète's formulas are derived? That's easy: you simply multiply out the factorization of the polynomial.

    Incidentally, the product of the roots can be done in essentially the same way as you did with the sum of all the roots in the even case.

    First off, e^z is usually used as shorthand for exp(z), which is not a multivalued function.

    However, if you really do intend it as complex exponentiation, then e^(ix) is in fact multivalued. (And really does have multiple values in the case that x is not purely imaginary)

    Remember, complex exponentiation is, by definition, given by the formula
    [tex]a^b = exp(b \log a)[/tex]
  4. Jan 9, 2009 #3
    Yes the combinatorials add up to zero because they are the coefficients of the polynomial x^n - 1 = 0 (eg. zero).
  5. Jan 9, 2009 #4


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    More simply, [itex](z- a_1)(z- a_2)\cdot\cdot\cdot(z- a_n)= z^n\cdot\cdot\cdot- (a_1+ a_2+\cdot + a_n)z+ (a_1a_2\cdot\cdot\cdot a_n)= z^n- 1[/itex]. Since the coefficient of z in [itex]z^n-1[/itex] is 0, so is the sum [itex]a_1+ a_2+\cdot\cdot\cdot +a_n= 0[/itex].
  6. Jan 10, 2009 #5

    Gib Z

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    Minor correction, but you started it from the wrong end, its the coefficient of z^(n-1) that matters here. =]
  7. Jan 10, 2009 #6
    Another way is to multiply the sum by one of the roots. This will rotate all roots by that roots argument, so you get back the same sum, ie:
    we can choose e^ix=/=1 (unless its the first root of unity, where the theorem fails anyway), so S=0.
  8. Jan 11, 2009 #7
    It seems to me very related to a regular polygon being a closed figure; which is what you get if you add these vectors geometrically. Though probably that is not really a proof.
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