Evaluating the complex summation

Question

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

See https://math.stackexchange.com/questions/2209034/finding-sum-k-0n-1-dfrac-alpha-k2-alpha-k-where-alpha-k-is-prim and related ones like https://math.stackexchange.com/questions/1811081/problem-based-on-sum-of-reciprocal-of-nth-roots-of-unity — lab bhattacharjee, Mar 14 '18 at 16:25
You know that $f(z) = z^n - 1 = (z-z_1)\cdots (z-z_{n})$ (with $z_n = 1$) and note that $f'(z)/f(z) = \sum \frac{1}{z-z_i}$ — Winther, Mar 14 '18 at 16:27

Jack D'Aurizio · Answer 1 · 2018-03-15T03:23:22.833

3

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 $\zeta_1=1$.

edited Mar 15 '18 at 03:23

answered Mar 14 '18 at 20:41

Jack D'Aurizio

353,855

What does that symbol before $k$ mean? How to read it? – Archer 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
1

How did you find that sum is $$\frac {nz^{n-1}}{z^n-1}$$ – Rohan Shinde Mar 15 '18 at 01:45
@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 03:24
Oh got that thanks (+1) – Rohan Shinde Mar 15 '18 at 04:56

CY Aries · Answer 2 · 2018-03-14T17:29:43.287

2

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

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

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

edited Mar 14 '18 at 17:29

answered Mar 14 '18 at 17:19

CY Aries

23,393

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 $y$? – Archer 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... – Archer Mar 14 '18 at 17:39
As in the question, $1,z_1,z_2,\dots,z_{n-1}$ are the roots of unity. $1$ is one of the roots of $z^n=1$. So naturally, $\frac{1}{3-1}$ is a root of $(3y-1)^n=y^n$. You may take $z_0=1$, then $\frac{1}{2}$ is just $y_0$. – CY Aries Mar 14 '18 at 17:42
Let us continue this discussion in chat. – Archer Mar 14 '18 at 17:57

Evaluating the complex summation

2 Answers2