Awesome Roots Of Unity

$P(z)$ can be expressed as follows:

$z^{n+1}-1=(z-1)(z-\alpha )(z-\alpha ^{2})(z-\alpha ^{3})...(z-\alpha ^{n})$

Taking log both sides:

$log(z^{n+1}-1)=log(z-1)+log(z-\alpha)+log(z-\alpha ^{2})+...log(z-\alpha ^{n})$

Differentiating both sides with respect to z:

$\frac{1}{z^{n+1}-1}.(n+1)z^{n}=\frac{1}{z-1}+\frac{1}{z-\alpha }+\frac{1}{z-\alpha^{2} }+...+\frac{1}{z-\alpha^{n} }$

$\Rightarrow \frac{1}{z^{n+1}-1}.(n+1)z^{n}-\frac{1}{z-1}=\frac{1}{z-\alpha }+\frac{1}{z-\alpha^{2} }+...+\frac{1}{z-\alpha^{n} }$

Replacing z by 'm' we get the required expression:

$\frac{1}{m-\alpha }+\frac{1}{m-\alpha^{2} }+...+\frac{1}{m-\alpha^{n} }= \frac{1}{m^{n+1}-1}.(n+1)m^{n}-\frac{1}{m-1}$

$=\frac{(n+1)m^{n}(m-1)-(m^{n+1}-1)}{(m^{n+1}-1)(m-1)}$

On rearranging the terms of numerator after opening brackets:

$\frac{1}{m-\alpha }+\frac{1}{m-\alpha^{2} }+...+\frac{1}{m-\alpha^{n} }=\boxed{\frac{m^{n}(mn-n-1)+1}{(m^{n+1}-1)(m-1)}}$

It's good to have you back, Compañero Alan... I love your problems!

Awesome roots of unity indeed! The problem turns out to be fairly straightforward as long as we keep our $m$ 's and $n$ 's straight ;)

Make a substitution $w=\frac{1}{m-z}$ or $z=\frac{mw-1}{w}$ . Now $z^{n+1}=1$ gives $(mw-1)^{n+1}=w^{n+1}$ or $(m^{n+1}-1)w^{n+1}-(n+1)m^nw^n+...=0$ . By Viète, the sum over all the roots of unity $\alpha^k$ for $0\leq k \leq n$ is $\frac{(n+1)m^n}{m^{n+1}-1}$ . Subtracting $\frac{1}{m-1}$ for $\alpha^0=1$ gives $\boxed{\frac{m^n(mn-n-1)+1}{(m^{n+1}-1)(m-1)}}$ .

Here we need to assume that $m$ isn't an $(n+1)^{th}$ root of unity.

Just consider value of n=2 it will convert into cube roots of unity and then solve

Assume $P(z) = z^{n} + z^{n-1} ....+z+1$ = $(x-\alpha)(x-\alpha^{2})....(x-\alpha^{n})$ . Take natural logarithm on both sides of equation , and $differentiate$ . Use sum of $n$ terms if an GP $S_n$ = $a[r^{n}- 1/r-1]$ , where $a$ is first term of a GP containing $n$ terms with a common ratio $r$ . Rearrange to get the desired answer.