$\omega = \cos{\left( \frac{2 \pi}{n} \right)} + \sin{\left( \frac{2 \pi}{n} \right)}i$, where $n$ is an integer greater than 1.

Please demonstrate the following without Euler's Formula.

  1. Show that $\omega^n = 1$.

This one should be a simple expansion of de moivres showing that the n's cancel out and then is equal to 1

  1. If $a$ and $b$ are integers such that $a \equiv b \bmod{n}$, prove that $\omega^a = \omega^b$.

Not sure what to do on this one

Consider the set $ \Omega_n := \{ 1, \omega,\dots,\omega^{n-1} \}. $

  1. Prove that $\Omega_n$ is closed with respect to complex multiplication, i.e., given $\alpha, \beta \in \Omega_n$, show that $\alpha \beta \in \Omega _n$.

Using problem 1 all items in set are equal to 1 so 1 times 1 is equal to 1.

  1. Given $\alpha \in \Omega_n$, prove that there exists $\beta \in \Omega_n$ such that $\alpha \beta = 1$.

since all $\alpha$ is equal to 1 then $\beta$ simply has to the the item $\{1\}$ from the set.

There are my pseudo proofs. I feel i have done number 3 and 4 wrong since my answer is so simple.

  • $\begingroup$ cant use Euler's $\endgroup$ – aeipownu Oct 13 '17 at 19:47
  • $\begingroup$ You should really ask one question at a time. $\endgroup$ – Morgan Rodgers Oct 13 '17 at 19:49
  • $\begingroup$ I guess so. Don't really know the standard. Figured they were small enough that if someone were to search for this subject all in one would be better. ill split them next time. $\endgroup$ – aeipownu Oct 13 '17 at 19:51

Apparently, based on the text of the question, de Moivre's theorem is permissible here. It is:

$(\cos \theta + i\sin \theta)^m =\cos m\theta +i \sin m\theta \tag 1$

where $m \in \Bbb N$; if we take

$\omega = \cos \dfrac{2\pi}{n} + i\sin \dfrac{2\pi}{n}, \tag 2$

then from equation (1), with $\theta = (2\pi/n)$,

$\omega^n = (\cos \dfrac{2\pi}{n} + i\sin \dfrac{2\pi}{n})^n = \cos (n \dfrac{2\pi}{n}) + i \sin (n \dfrac{2\pi}{n})$ $=\cos 2\pi +i \sin 2\pi = \cos 2\pi = 1; \tag 3$

that's item (1).

As for item (2), if

$a \equiv b \pmod n, \tag 4$

then $a = b + kn$ for some $k \in \Bbb Z$; thus

$\omega^a = \omega^{b + kn} = \omega^b (\omega^n)^k = \omega^b (1)^k = \omega^b; \tag 5$

for item (3), if $\alpha, \beta \in \Omega_n$, we may take

$\alpha = \omega^i, \; \beta = \omega^j \tag 6$

with $0 \le i, j \le n -1$; if $i + j \le n - 1$, then

$\alpha \beta = \omega^i \omega^j= \omega^{i + j} \in \Omega_n; \tag 7$

if $i + j \ge n$ then we have $n \le i + j \le 2n - 2$ and so $i + j = n + k$ where $0 \le k \le n - 2$, and

$\alpha \beta = \omega^i \omega^j= \omega^{i + j} = \omega^{n + k} = \omega^n\omega^k = \omega^k \in \Omega_n, \tag 8$

via (3); finally, as for item (4), with

$\alpha = \omega^i, \; 0 \le i \le n - 1, \tag 9$

we have $1 \le n - i \le n$; if $n - i < n$, then

$\beta = \omega^{n - i} \in \Omega_n \tag{10}$


$\alpha \beta = \omega^i \omega^{n - i} = \omega^n = 1, \tag{11}$

whereas if $n - i = n$, i.e. if $i = 0$, then $\alpha^i = \alpha^0 = 1$ and we have $\beta = \omega^{n - i} = \omega^n = 1$ and $\alpha \beta = 1$.

We have demonstrated items (1)-(4), as per request.

  • $\begingroup$ in line 6 do you mean beta is $\omega^j$? $\endgroup$ – aeipownu Oct 14 '17 at 23:45
  • $\begingroup$ @aeipownu: indeed; fixed; thanks! $\endgroup$ – Robert Lewis Oct 15 '17 at 0:15

I think that you are misunderstanding a couple of the questions.

For question 1 I think what you have should do it. The important thing is to remember that after question 1, you have some fixed $n$. Say (for arguments sake) that your $n$ is $5$.

For question 2, I recommend that you think about what the statement actually means. It means $a = b + kn$ for $k \in \mathbb{Z}$. Remember that your $n$ is fixed. What you really have is that $a$ and $b$ differ by some multiple of $n$. From what I can see, expanding out what you are given should reap rewards.

For question 3, remember that $n$ is fixed. If $\omega^n = \omega^5 = 1$, that does not mean that $\omega^{n-1} = 1$. What I mean to say is that your assertion that all members of that set are equal to $1$ is incorrect. You need to show that multiplying any two members of that set together brings you another member of the set.

For question 4 you have made the same false assertion, and your idea does not work. You need to show that for any arbitrary element of that set, there will always be another element which when multiplied by the first element gives you 1. This is intuitively true, but you have to prove it rigorously.

  • $\begingroup$ 3. lets say I put $\omega$ to the power of integers x and y. When I expand it out using demoivres they will simply cancel out and I will end up with the original. Then it's just like squaring and I would use complex multiplication then show that that becomes the original which is equal to 1? $\endgroup$ – aeipownu Oct 13 '17 at 19:36


$a \equiv b\pmod n \implies a = b + qn$

$\omega^a = \omega^{b+qn} = (\omega^b)(\omega^{qn}) = (\omega^b)(\omega^{n})^q =(\omega^b)(1)^q =\omega^b$


$\omega^a = (cos \frac {2a\pi}{n} + i\sin \frac {2a\pi}{n})$

$\omega^a\omega^b = (cos \frac {2a\pi}{n} + i\sin \frac {2a\pi}{n}) (cos \frac {2b\pi}{n} + i\sin \frac {2b\pi}{n}) = (cos \frac {2(a+b)\pi}{n} + i\sin \frac {2(a+b)\pi}{n}) = \omega^{a+b}$

if $a+b > n, \exists c \in \{1,\cdots n-1\} : a+b \equiv c \pmod n$

and $\omega^a\omega^b = \omega^c$


$\omega^a\omega^{n-a} = 1$

  • $\begingroup$ 4. What if your alpha is 1? Edit: 1*1=1 so never-mind on that one. $\endgroup$ – aeipownu Oct 13 '17 at 19:41

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