Complex numbers

**PacManisAlive** · July 3rd 2007, 07:04 AM

I am trying to solve this set of problems, any ideas:

Let 1, omega, omega squared, ....., omega^n-1 be nth rooots of unity
then
(a) Show the conjugate of any nth root of unity is another root of unity, by expressing omega(bar)^j in the form omega^k for appropriate k.
(b) Find the product of the nth roots of unity
(c) Find the sum Summation (n-1), k=0 for omega^k of nth roots of unity. (this is a geometric series).

**ThePerfectHacker** · July 3rd 2007, 08:46 AM

Originally Posted by PacManisAlive

I am trying to solve this set of problems, any ideas:

Let 1, omega, omega squared, ....., omega^n-1 be nth rooots of unity

Let $\zeta$ to the generator of the $n$ roots of unity. That means, $\{1,\zeta,\zeta^2,...,\zeta^{n-1}\}$ contains the roots of unity.

(a) Show the conjugate of any nth root of unity is another root of unity, by expressing omega(bar)^j in the form omega^k for appropriate k.

Choose $j=n-k$ .
Then,
$\cos \left( \frac{2\pi (n-k)}{n} \right) + i \sin \left( \frac{2\pi (n-k)}{n} \right) = \cos \left( \frac{2\pi k}{n}\right) - i \sin \left( \frac{2\pi k}{n} \right)$

(b) Find the product of the nth roots of unity

We have, (since $\zeta \not = 1$ ).
$1+\zeta + \zeta^2 + \zeta^3 + ... + \zeta^{n-1} = \frac{1-\zeta^n}{1-\zeta} = 0$
Because $\zeta^n - 1=0$ .

(c) Find the sum Summation (n-1), k=0 for omega^k of nth roots of unity. (this is a geometric series).

How is that different from (b)?

**Soroban** · July 3rd 2007, 08:58 AM

Hello, PacManisAlive!

Here's some help . . .

Let $1,\:\omega,\:\omega^2,\:\omega^3,\:\cdots\,\:\omeg a^{n-1}$ be the $n^{th}$ roots of unity.

(a) Show the conjugate of any $n^{th}$ root of unity is another root of unity.

Let a root be: . $a + bi \;=\;\cos\theta + i\sin\theta$

Then: . $(a + bi)^n \;=\;(\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta) \;=\;1$

. . Hence: . $\begin{array}{ccc}\cos(n\theta) & = & 1 \\ \sin(n\theta) & = & 0 \end{array}$

The conjugate is: . $a - bi \;=\;\cos(\text{-}\theta) + i\sin(\text{-}\theta) \;=\;\cos\theta - i\sin\theta$

$\text{Then: }\;(a - bi)^n \;=\;(\cos\theta - i\sin\theta)^n \;=\;\underbrace{\cos(n\theta)}_{\text{this is 1}} - i\underbrace{\sin(n\theta)}_{\text{this is 0}} \;=\;1$

Therefore, the conjugate is also an $n^{th}$ root of unity.

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