1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit help needed for end of complex number question

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    43. Let ##w_1, w_2, ... , w_n## be the ##n## distinct ##n##'th roots of unity ##(n\geq0)##. Show that if ##k## is an integer then $$w_1^k+w_2^k+...+w_n^k$$ equals ##0## or ##n##. Find the values of ##k## for which the sum is ##n##.

    Hint:Write the roots in polar form and sum the resulting geometric progression.

    2. Relevant equations
    It has been shown elsewhere that $$1-e^{i\theta}=e^{i\frac{\theta}{2}}.-2i.\sin{\left(\frac{\theta}{2}\right)}$$
    which has been derived from Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta.$$

    3. The attempt at a solution
    Note: My problems start with Case 2 at the end of my solution. I have questions after the end of my solution.

    Solution:

    In polar form the ##n##'th roots of unity can be written as $$w_l=e^{i\pi\frac{(2l-n)}{n}}$$ for ##l=1## to ##n##.

    Let the sum in the question be ##S_n## then this can be written in polar form as
    $$S_n=\left(e^{i\pi\frac{(2.1-n)}{n}}\right)^k+\left(e^{i\pi\frac{(2.2-n)}{n}}\right)^k+...\left(e^{i\pi\frac{(2.n-n)}{n}}\right)^k.$$ There is a common factor of $$e^{i\pi\frac{(2-n)k}{n}}$$ so we can take this out side brackets to get
    $$S_n=e^{i\pi\frac{(2-n)k}{n}}
    \left(1+e^{i\pi\frac{2k}{n}}+...+e^{i\pi\frac{2(n-1)k}{n}}\right).$$
    We have a geometric progression in the brackets with a ratio of ##e^{i\pi\frac{2k}{n}}## so we can now write the sum as
    $$S_n=e^{i\pi\frac{(2-n)k}{n}}.\frac{1-\left(e^{i\pi\frac{2k}{n}}\right)^n}{1-\left(e^{i\pi\frac{2k}{n}}\right)}
    =e^{i\pi\frac{(2-n)k}{n}}.\frac{1-e^{i\pi2k}}{1-e^{i\pi\frac{2k}{n}}}.$$

    As $$1-e^{i\theta}=e^{i\frac{\theta}{2}}.-2i.\sin{\left(\frac{\theta}{2}\right)}$$ the sum can be written as
    $$S_n=e^{i\pi\frac{(2-n)k}{n}}.
    \frac{e^{i\pi k}.-2i.\sin{(\pi k)}} {e^{\frac{i\pi k}{n}}.-2i.\sin{(\frac{\pi k}{n})}}
    =e^{i\pi\frac{k}{n}}.\frac{\sin{(\pi k)}} {\sin{(\pi\frac{k}{n})}}.$$

    I now split up the solution into two cases.

    Case 1 - The integer ##k## is not an integer multiple of ##n##.

    In this case ##\sin{(\pi k)}=0## (it always will as ##k## in an integer), ##\sin{(\pi\frac{k}{n})}\neq0## as ##n## is not a factor of ##k## and ##\frac{k}{n}## does not reduce to an integer, so ##S_n=0##.

    Case 2 - The integer ##k## is an integer multiple of ##n##.

    Let ##k=qn## where ##q\in\mathbb Z##. The problem here is that both ##\sin{(\pi k)}## and ##\sin{(\pi\frac{k}{n})}## equal zero, so we have the indeterminate form ##\frac{0}{0}## in ##S_n##.

    You can check the limit of ##S_n## as ##q\rightarrow0##
    $$\lim_{q\rightarrow0}\left(e^{i\pi\frac{k}{n}}. \frac{\sin{(\pi k)}} {\sin{(\pi\frac{k}{n})}}\right)=
    \lim_{q\rightarrow0}\left(e^{i\pi\frac{qn}{n}}. \frac{\sin{(\pi qn)}} {\sin{(\pi\frac{qn}{n})}}\right)=
    \lim_{q\rightarrow0} \left(e^{i\pi q}. \frac{\sin{(\pi qn)}} {\sin{(\pi q)}}\right),$$
    and then
    $$=\lim_{q\rightarrow0}e^{i\pi q}.
    \lim_{q\rightarrow0} \left(\frac{\sin{(\pi qn)}} {\pi qn} \right).
    \lim_{q\rightarrow0} \left(\frac{\pi q}{\sin{(\pi q)}}\right).
    \lim_{q\rightarrow0} n =
    1.1.1.n=n\qquad.\Box$$

    This is as far as I have got. My problems now are
    1) How do I prove Case 2 for the general situation when ##q## is any integer?
    2) How do I show that the limit in Case 2 is continuous?
    3) How so I deal with the fact that ##e^{i\pi q}## flips between ##\pm1## depending if ##q## is odd or even? I guess question 3) here will be answered by 1) and 2) depending on the sign of the sin limits.

    Thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 30, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi EdMel! :smile:

    patient to doctor: doctor, it hurts when i do that

    doctor to patient: then don't do that! :biggrin:
    so don't use Sn :wink:
     
  4. May 31, 2012 #3
    Hi tiny-tim,

    I can see that for Case 2 its easer to use $$S_n=e^{i\pi\frac{(2-n)k}{n}}\left(1+e^{i\pi\frac{2k}{n}}+...+e^{i\pi \frac{2(n-1)k}{n}}\right)$$ instead of reducing this to include the sin functions. Substituting in ##k=nq## gives
    $$S_n=e^{i\pi\frac{(2-n)nq}{n}}\left(1+e^{i\pi\frac{2nq}{n}}+...+e^{i\pi \frac{2(n-1)nq}{n}}\right)=e^{i\pi2q}.e^{i\pi(-nq)}\left(1+e^{i\pi2q}+...+e^{i\pi2(n-1)q}\right),$$
    all the ##e^{i\theta}## terms with an even integer multiple of ##\pi## as an argument are equal to 1
    $$S_n=1.e^{i\pi(-nq)}(1+1+...+1)=e^{i\pi(-nq)}.n\qquad,$$
    which still has ##S_n## flipping between ##-n## and ##n## depending on ##k## being and odd or even integer multiple of ##n##.

    Maybe my algebra was wrong some-where? I have checked it several times though.

    Thanks,

    Ed
     
  5. May 31, 2012 #4
    I just realized I'm over complicating things.

    From the question and with ##k=nq## we have
    $$w_1^k+w_2^k+...+w_n^k=w_1^{nq}+w_2^{nq}+...+w_n^{nq}=(w_1^n)^q+(w_2^n)^q+...+(w_n^n)^q=(1)^q+(1)^q+...+(1)^q=n.\Box$$

    In my original solution I have subtracted ##\pi## from the arguments of the n'th roots so they fit in the range ##(-\pi,\pi]## and I think this has stuffed me up.
     
  6. May 31, 2012 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit help needed for end of complex number question
Loading...