- #1
EdMel
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Homework Statement
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.
Homework 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 $$e^{i\theta}=\cos\theta+i\sin\theta.$$
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.