Telescoping with imaginary numbers If $\omega^{1997} = 1$ and $\omega \neq 1$, then
$$ \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}}$$
can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.
I really can't seem to find the complex number w that satisfies this condition, and I cannot find any patterns/telescoping methods.  Can anyone help me or give me some pointers?
Thanks!
 A: Hint: Just apply the old Gauss trick ! We have 

$$\dfrac1{1+w^k}~+~\dfrac1{1+w^{1997-k}}~=~\dfrac1{1+w^k}~+~\dfrac{w^k}{w^k+w^{1997}}~=~\dfrac1{1+w^k}+\dfrac{w^k}{w^k+1}=1.$$

If the sum would have started with $~\dfrac1{1+w^0}=\dfrac12~,~$ then the final result would have been 
$\dfrac{1998}2~.~$ As it stands, however, we must subtract $~\dfrac12,~$ thus arriving at $~\dfrac{1997}2~.$
A: Hint:  Note that $x=\omega^k$ for $k=1,2,\ldots,1996$ are all the roots of $1+x+x^2+\ldots+x^{1996}=0$, whence they satisfy the equality
$$\frac1{1+x}=-\frac{x+x^2+\ldots+x^{1995}+x^{1996}}{1+x}=-x-x^3-\ldots-x^{1995}\,.$$
If you replace $1997$ by any odd natural number $p$, then the required sum is $\frac{p}2$.  
If $t$ is an arbitrary complex number, $p$ is a natural number (not necessarily odd), and $\omega$ is a primitive $p$-th root of unity, then one can also show that $$\sum_{k=0}^{p-1}\,\frac{1}{1-t\omega^k}=\frac{p}{1-t^p}\,$$ provided that $t$ is not a $p$-th root of unity.  This can also be seen as an identity in $\mathbb{C}(t)$, or even in $K(t)$ where $K$ is a field whose characteristic does not divide $p$ and which contains a splitting field of the polynomial $t^p-1$.
A: Clearly, $w^r,1\le r\le n;$ are the roots of $$x^n=1$$
Set $\dfrac1{1+x}=y\iff x=\dfrac{1-y}y$
$$\implies\left(\dfrac{1-y}y\right)^n=1$$
If $n$ is odd$=2m+1$(say), $$2y^{2m+1}-\binom {2m+1}1y^{2m}+\cdots=0$$
Using Vieta's formula,
$$\sum_{r=1}^{2m+1}\dfrac1{1+w^r}=\dfrac{2m+1}2$$
A: We have $\omega^N=1$.  For odd $N$, the roots are 
$$\omega_{n} =e^{i2n \pi/N} \tag 1$$
for $n =0,\pm1,\pm2,\cdots, \pm(N-1)/2$.  Then, we have
$$\begin{align}
\sum_{n=-(N-1)/2}^{(N-1)/2}\dfrac{1}{1+\omega_{n}}&=\sum_{n=-(N-1)/2}^{(N-1)/2}\dfrac{1}{1+e^{i2n \pi/N}} \tag 2\\\\
&=\dfrac12+2\text{Re}\left(\sum_{n=1}^{(N-1)/2}\dfrac{1}{1+e^{i2n \pi/N}}\right)\tag 3\\\\
&=\dfrac12+2\text{Re}\left(\sum_{n=1}^{(N-1)/2}\dfrac{e^{-in\pi/N}}{2\cos(n\pi/N)}\right) \tag 4\\\\
&=\dfrac12+\dfrac{N-1}{2} \tag 5\\\\
&=\dfrac N2
\end{align}$$
For $N=1997$, the remainder of $\dfrac{N+2}{1000}=\dfrac{1999}{1000}$ is $999$.
NOTES:
In arriving at $(2)$, we made use of $(1)$, which provided the roots of unity for $\omega^N-1=0$.
In going from $(2)$ to $(3)$, we used the facts that (i) $\dfrac{1}{1+\omega_0}=\dfrac12$ and (ii) the roots are complex conjugates with $\omega_{-n}=\bar \omega_{n}$, where $\bar \omega_{n}$ denotes the complex conjugate of $\omega_{n}$.  
In addition, we used the identity that $z+\bar z=2\text{Re}(z)$ for complex numbers.  Therefore, we recognized that $\dfrac{1}{1+\omega_n} +\dfrac{1}{1+\bar\omega_n}=2\text{Re}\left(\dfrac{1}{1+\omega_n}\right)$.
In going from $(3)$ to $(4)$, we multiplied the summands by $1=\dfrac{e^{-in\pi/N}}{e^{-in\pi/N}}$ and used Euler's Identity in the form $e^{iz}+e^{iz}=2\cos z$.  Thus, $\dfrac{1}{1+e^{i2n\pi/N}}=\dfrac{e^{-in\pi/N}}{2\cos (n\pi/N)}$.
Finally, in going from $(4)$ to $(5)$, we used Euler's Identity once more to write $\dfrac{e^{-in\pi/N}}{2\cos (n\pi/N)}=$ $=\dfrac{\cos (n\pi/N)-i\sin (n\pi/N)}{\cos (n\pi/N)}$, which after taking the real part becomes $1$.
