Limit help needed for end of complex number question

[EdMel]

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.

 

[tiny-tim]

Hi EdMel!

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

doctor to patient: then don't do that! ​

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

Let $k=qn$ … we have the indeterminate form $\frac{0}{0}$ in $S_n$.

so don't use S_n

 

[EdMel]

Hi EdMel!

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

doctor to patient: then don't do that! ​

so don't use S_n

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

 

[EdMel]

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

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.

 

[tiny-tim]