Complex roots of unity arrangement. The following question is from an entrance test named IAT (Indian Institute of Science and Education Research Aptitude test).

Let $1, z_2, z_3,\ldots, z_n$ be the roots of the equation: $$x^n-1=0,\quad n\ge3$$
Find the value of: $$\frac{1}{2-z_2}+\frac{1}{2-z_3}+\cdots+\frac{1}{2-z_n}$$

I have tried to use the formula
$$
\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+\ldots+1
$$
but failed. My experience with competitive exam problems suggests that there should be some trick which at the current stage I am unable to identify. I will be grateful to anyone who can provide a partial or a complete solution to the above problem.
 A: First of all, we have that the sum of all $n$th roots of unity is $0$, and also that the sum of all $n$th roots of unity to any (non-zero) integer power is also $0$ - I prove this at the end. I will be adding also the term: $\frac{1}{2-1}$ for reasons that will become clear soon - so just subtract $1$ from my answer. If we add all the fractions together, we must of course make the base the same. The resulting product on the denominator is:
$$\prod_{k=1}^n(2-z_k)$$
But a polynomial is the product of factors in its roots! This is just our polynomial $x^{n}-1$, evaluated at $x=2$. Hence the product is $2^n-1$ - this was why I included the root of $1$, to get these properties.
Now we need to sort out the numerator - this is harder. The $k$th numerator will be the product of all factors, divided by $(2-z_k)$. We have:
$$\frac{2^n-1}{2-z_k}=\frac{2^n}{2}\cdot\frac{1-\left(\frac{z_k}{2}\right)^n}{1-\frac{z_k}{2}}=2^{n-1}\cdot\sum_{m=0}^{n-1}\left(\frac{z_k}{2}\right)^m$$
Summing all the numerators, to obtain a final answer, we have:
$$2^{n-1}\cdot\sum_{m=0}^{n-1}\left(\frac{z_1}{2}\right)^m+2^{n-1}\cdot\sum_{m=0}^{n-1}\left(\frac{z_2}{2}\right)^m+2^{n-1}\cdot\sum_{m=0}^{n-1}\left(\frac{z_3}{2}\right)^m+\cdots$$
Which is:
$$2^{n-1}\sum_{m=0}^{n-1}\frac{1}{2^m}\sum_{k=1}^nz_k^m=2^{n-1}\left(n+\sum_{m=1}^{n-1}0\right)=n\cdot2^{n-1}$$
So the sum of all the fractions, including $\frac{1}{2-1}$, is:
$$\frac{n\cdot2^{n-1}}{2^n-1}$$
And the answer to your specific problem, minus the extra term of $1$ that I added, is:
$$\frac{n\cdot2^{n-1}}{2^n-1}-1$$
The proof that the sum of integer (non-zero) powers of the $n$th roots of unity is $0$:
The $k$th root of unity, of order $n$, is expressible as (I will index from $0-(n-1)$ for the sake of this proof): $$z_k=\exp\left(\frac{2\pi i}{n}\cdot k\right)=\exp\left(\frac{2\pi i}{n}\right)^k$$ If I take $m$ as a non-zero integer, I have: $$z_k^m=\exp\left(2\pi i\cdot\frac{m}{n}\right)^k$$ And so their sum can be given by the geometric series:
$$\sum_{k=0}^{n-1}z_k^m=\sum_{k=0}^{n-1}\exp\left(2\pi i\cdot\frac{m}{n}\right)^k=\frac{1-\exp\left(2\pi i\cdot\frac{m}{n}\right)^n}{1-\exp\left(2\pi i\cdot\frac{m}{n}\right)}=\frac{1-1}{1-\exp\left(2\pi i\cdot\frac{m}{n}\right)}\\=0,\quad m(\in\Bbb Z)\neq0\quad\text{mod $n$}$$
