$1,z_1,z_2,z_3,...z_{n-1}$ are the $n^{\text{th}}$ roots of unity, then the value of $\dfrac 1{3-z_1}+ \dfrac{1}{3-z_2}+...+\dfrac 1 {3-z_{n-1}}$ is equal to?
I wrote the polar form of the $n$th root of unity $(\cos \dfrac{2k\pi}{n}+i \sin\dfrac{ 2k\pi}{n})$ $\forall ~ k\in\{0,1,2...n-1\} $
But that didn't help at all. How do I go about solving this problem?
On
If a polynomial $q(z)$ is such that $$ q(z)=\prod_{k=1}^{n}(z-\zeta_k) $$ then by applying $\frac{d}{dz}\log(\cdot)$ to both sides we get $$ \frac{q'(z)}{q(z)}=\sum_{k=1}^{n}\frac{1}{z-\zeta_k} $$ hence your sum is just $\frac{n z^{n-1}}{z^n-1}$ evaluated at $z=3$, i.e. ${\frac{n 3^{n-1}}{3^n-1}}$, minus the contribution due to $\zeta_1=1$.
For $k=1,2,\dots n-1$, $z_k^n=1$. Let $\displaystyle y_k=\frac{1}{3-z_k}$.
$$\left(3-\frac{1}{y_k}\right)^n=z_k^n=1$$
$$(3y_k-1)^n=y_k^n$$
$\displaystyle \frac{1}{2}, y_1,y_2,\dots,y_{n-1}$ are the roots of $(3y-1)^n-y^n=0$.
$$\frac{1}{2}+\sum_{k=1}^{n-1}y_k=\frac{n\cdot 3^{n-1}}{3^n-1}$$
$$\sum_{k=1}^{n-1}y_k=\frac{n\cdot 3^{n-1}}{3^n-1}-\frac{1}{2}$$