Complex Numbers: Solving 2^(8n)exp^(2niπ)=2^(8n)

[craig100]

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

I've come across 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

 

[cristo]

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

 

[Eighty]

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

 

[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]

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

 

[Gib Z]

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

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

 

[craig100]

Thank guys, quite simple afterall