Complex number roots of unity proof

Question

Suppose $w_1$, $w_2$, ..., $w_{n-1}$ are the complex roots not equal to $1$ of $z^n-1=0$, where $n$ is odd.

Show: $\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}$.

Hello, I am struggling with this complex number proof. Any help is welcomed; thank you in advance.

I have tried expanding the LHS and comparing it to the RHS but am not able to make them equal; I got this far: $\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}$

Part 2 of the question: Hence show: $$\sum_{k=1}^{n-1} \frac{1-\bar{w_k}}{1+w_k} = n$$

Any suggestions?

Answer 1

Since $n$ is odd we know: $w\ne-1$. Therefore, using $(1+w)(1+\bar w)\ne0$ and $w\bar w=1$: $$\begin{align} &\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}\\ \iff&\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}\\ \iff&2-w^2-\bar{w}^2=2-w^2-\bar{w}^2, \end{align}$$ the latter equation being obvious identity.

Answer 2

The idea is that any root of unity complies $w\bar{w}=1$ and the main manipulation trick you need is: $$ (w+\bar{w})^2 = w^2 + 2w\bar{w} + \bar{w}^2 = 2 + w^2+\bar{w}^2 $$ so that $2-w^2-\bar{w}^2 = 2 - (w^2+\bar{w}^2) = 2 - ( (w+\bar{w})^2-2 ) = 2^2-(w+\bar{w})^2 $ which is a difference of squares. Can you fill the remaining details?

Answer 3

From $w^n=1$ we deduce $\bar w^n=1$ and $(w\bar w)^n=1$. Then as $w\bar w$ is real and $n$ is odd, $w\bar w=\sqrt[n]1=1$.

The given identity holds for $w=1$. Now assuming $w\ne1$, we multiply both sides of the equation by $(1+w)(1+\bar w)=2-w-\bar w\ne0$ and get

$$2-w^2-\bar w^2=4-w^2-2w\bar w-\bar w^2,$$ which completes the proof.

Answer 4

Since $\omega$ lies on the unit circle, more generally, let $\omega = \cos \theta + i\sin\theta$ and $\overline{\omega} = \cos \theta-i\sin\theta$:

$$\text{LHS} = \frac{2-\omega^2-\overline{\omega}^2}{2+\omega+\overline{\omega}}$$ $$= \frac{2 - (\cos^2 \theta - \sin^2 \theta + 2i \cos \theta \sin \theta)-(\cos^2 \theta - \sin^2 \theta- 2i\cos \theta \sin \theta)}{2 + 2\cos\theta}$$ $$= \frac{2 + 2\sin^2 \theta- 2 \cos^2 \theta}{2 + 2 \cos \theta} = \frac{1 + (1 - \cos^2 \theta) - \cos^2 \theta}{1 + \cos \theta} = \frac{2(1 + \cos \theta)(1 - \cos \theta)}{1 + \cos \theta} = 2 - 2 \cos \theta$$ $$= 2 - (\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta) = 2 - \omega - \overline{\omega} = \text{RHS}$$