Can't get enough of those Roots of Unity

$\displaystyle \sum_{k=1}^4 \frac{1}{1+\omega^k + \omega^{2k}} \\ \displaystyle = \frac{1}{1+\omega + \omega^2} + \frac{1}{1+\omega^2 + \omega^4} + \frac{1}{1+\omega^3 + \color{#3D99F6}{\omega^6}} + \frac{1}{1+\omega^4 + \color{#3D99F6}{\omega^8}} \quad \quad \small \color{#3D99F6} {[\omega^5 = 1]}\\ \displaystyle = \frac{1}{\color{#D61F06}{1+\omega + \omega^2}} + \frac{1}{\color{#D61F06}{1+\omega^2 + \omega^4}} + \frac{1}{\color{#D61F06}{1+\omega^3 + }\color{#3D99F6}{\omega}} + \frac{1}{\color{#D61F06}{1+\omega^4 +} \color{#3D99F6}{\omega^3}} \\ \displaystyle = \color{#D61F06}{-}\frac{1}{\color{#D61F06}{\omega^3 + \omega^4}} \color{#D61F06}{-} \frac{1}{\color{#D61F06}{\omega + \omega^3} } \color{#D61F06}{-} \frac{1}{\color{#D61F06}{\omega^2 + \omega^4}} \color{#D61F06}{-} \frac{1}{\color{#D61F06}{\omega + \omega^2}} \quad \quad \small \color{#D61F06}{ [1+\omega + \omega^2 +\omega^3 + \omega^4 = 0]} \\ \displaystyle = -\frac{1}{\omega^3(1 + \omega)} - \frac{1}{\omega(1 + \omega^2)} - \frac{1}{\omega^2(1 + \omega^2)} - \frac{1}{\omega(1 + \omega)} \\ \displaystyle = -\frac{\omega^2+\omega^4}{1 + \omega} - \frac{\omega^4+\omega^3}{1 + \omega^2} \\ \displaystyle = \frac{1+\omega+\omega^3}{1 + \omega} + \frac{1+\omega+\omega^2}{1 + \omega^2} \\ \displaystyle = 1 + \frac{\omega^3}{1 + \omega} + 1 + \frac{\omega}{1 + \omega^2} \\ \displaystyle = 2 + \frac{\omega^3(1 + \omega^2)+\omega(1 + \omega)}{(1 + \omega)(1 + \omega^2)} \\ \displaystyle = 2 + \frac{\omega^3 + \omega^5+\omega + \omega^2}{1 + \omega + \omega^2 + \omega^3} \\ = \boxed{3}$

This was really fun to solve. @Otto Bretscher , I don't know if you saw what I mentioned in my note yesterday, but your problem, one more of those , was a question in yesterday's paper and I solved it pretty quickly. Thank you!

The solution to this problem is based on the repeated application of this identity:

$1+\omega+\omega^2+\omega^3+\omega^4=0.$

So

$1+\omega+\omega^2=- \omega^3 - \omega^4\\ 1+\omega^2+\omega^4=-\omega-\omega^3\\ 1+\omega^3+\omega^6=1+\omega^3+\omega=-\omega^2-\omega^4\\ 1+\omega^4+\omega^8=1+\omega^4+\omega^3=-\omega-\omega^2$

If you apply these identities, you should get:

$\frac { -1 }{ \omega ^{ 2 } } (\frac {1 }{ \omega (\omega+1) } +\frac{1}{(\omega^2+1)}) \\-\frac{1}{\omega}(\frac{1}{\omega^2+1}+\frac{1}{\omega+1}).\\$

Taking LCM and simplifying the numerator (again using the identities mentioned above) we get

$-\frac { \omega ^{ 2 }+\omega +2 }{ \omega (\omega +1)(\omega ^{ 2 }+1) } \\ -\frac { 1 }{ \omega ^{ 2 } } (\frac { \omega ^{ 2 }-\omega ^{ 3 }-\omega ^{ 4 } }{ \omega (\omega ^{ 2 }+1)(\omega +1) } ).\\ \text{ Note that }(\omega ^{ 2 }+1)(\omega +1)=-\omega ^{ 4 }.\\ \text{So the denominator gets simplified to -1 = }-\omega^4\times\omega$

Now we get:

$2+\omega+\omega^2+1-\omega-\omega^2=\boxed 3.$

In general, for $k$ non-negative integer, when $w=e^{\frac{2\pi i}{3k+2}}$ , that is, when $w$ is the "smallest-angle" non-trivial $(3k+2)\mbox{-}th$ root of unity, the result of the sum is $2k+1$ .

In fact the result holds for $w$ being any non-trivial $(3k+2)\mbox{-}th$ root of unity.

Another possibility: Expand by $1+w^{2k}$ to simplify the denominator:

$\begin{aligned} (1+w^k+w^{2k})(1+w^{2k})&=w^{2k}+\sum_{k=0}^4 w^k=w^{2k}\\\\ \Rightarrow \quad\sum_{k=1}^4\frac{1}{1+w^k+w^{2k}}&=\sum_{k=1}^4\frac{1+w^{2k}}{w^{2k}}=4+\sum_{k=1}^4w^{-2k}=3+\sum_{k=0}^4w^{-2k}=3+0 \end{aligned}$

Rem.: In the last step, we use that $w^{-2}\neq1$ is another root of unity of 5