# Evaluating the complex summation

$1,z_1,z_2,z_3,...z_{n-1}$ 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?

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 $\zeta_1=1$.

• What does that symbol before $k$ mean? How to read it? – Abcd Mar 14 '18 at 20:47
• @Abcd its a greek alphabet zeta. Its nothing but root of $q$ here. – King Tut Mar 14 '18 at 20:52
• How did you find that sum is $$\frac {nz^{n-1}}{z^n-1}$$ – Darkrai Mar 15 '18 at 1:45
• @Manthanein: $z^n-1$ is the polynomial vanishing at the $n$-th roots of unity and $\frac{d}{dz}(z^n-1)=n z^{n-1}$. – Jack D'Aurizio Mar 15 '18 at 3:24
• Oh got that thanks (+1) – Darkrai Mar 15 '18 at 4:56

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 $(3y-1)^n-y^n=0$.

$$\frac{1}{2}+\sum_{k=1}^{n-1}y_k=\frac{n\cdot 3^{n-1}}{3^n-1}$$

$$\sum_{k=1}^{n-1}y_k=\frac{n\cdot 3^{n-1}}{3^n-1}-\frac{1}{2}$$

• It's a typo. It should be $(3y_k-1)^n=y_k^n$. It comes from $y_k^n(3-\frac{1}{y_k})^n=y_k^n$ – CY Aries Mar 14 '18 at 17:30
• How did you get 1/2 as a root and how did you transform $y_k$ to $y$? – Abcd Mar 14 '18 at 17:34
• $1$ is a $n$-th root of unity. So $y_0=\frac{1}{3-1}$ should also be a root of $(3y-1)^n=y^n$. – CY Aries Mar 14 '18 at 17:35
• If you want to check whether $y_k$ is a root of $(3y-1)^n=y^n$, you just substitute $y$ by $y_k$ to see whether the equality holds. Of course, it does, as $(3y_k-1)^n=y_k^n$.for all $k$. – CY Aries Mar 14 '18 at 17:38
• I don't get why you included $1/2$ explicitly... – Abcd Mar 14 '18 at 17:39