Complex Numbers(2)

1. Apr 13, 2007

craig100

Hey again;
It doesn't seem to be my day for spotting the obvious;

I've come accross this problem now:

$$2^{8n}\exp^{2ni\Pi} = 2^{8n}$$

This imples that $$exp^{2ni\Pi} = 1$$, i know that $$exp^{i\Pi} =-1$$, but is that related?

hmm, Latex issues...i'll try and fix them :) - sorted
Craig

Last edited: Apr 13, 2007
2. Apr 13, 2007

cristo

Staff Emeritus
Well, using the exponential form: $$e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1$$. So, $$e^{2\pi ni}=\cos(2n\pi)+i\sin(2n\pi)$$ and since cos(2n pi)=cos(2 pi)=1, and sin(2n pi)=sin(2 pi)=0, the result exp(2ni pi)=1 is obtained

3. Apr 13, 2007

Eighty

The relation to that famous identity is
$$e^{2i\pi}=(e^{i\pi})^2=(-1)^2=1.$$

4. Apr 13, 2007

Edgardo

Hi craig100,

you can visualize the complex number $$e^{i \theta}$$ as a vector of length 1 that is rotated by an angle $$\theta$$ with respect to the positive x-axis,
see [URL [Broken].[/URL]

Also, see the section "Polar Form" here

The form $$z=r \cdot e^{i \theta}$$ is also called polar form of the complex number z. The vector has the length r and an angle $$\theta$$ with respect to the positive x axis.

If you visualize the complex number in that way, you can immediately see what value the complex number z will have for certain values such as
$$\theta = \pi/4, \pi/2, \pi, 2 \pi$$.

This example shows $$z=5 \cdot e^{i \pi /3}$$

Last edited by a moderator: May 2, 2017
5. Apr 15, 2007

Gib Z

Best way to show how its related is Eighty's post.

6. Apr 15, 2007

craig100

Thank guys, quite simple afterall