Euler's formula and its connection to the De Moivre's formula

[rtk]

first of all i'm sorry about my english.
i'm learning computer science and now im studing about the FFT algorithm which using the comlex unitiy roots, and i came to a strange conclusion.
if
e^((2*PI*k*i)/n) = cos(2*PI*k/n) + i*sin(2*PI*k/n)

[for i imaginary number, and k,n natural numbers]

for all k,n (from the Euler's formula), than why

(e^2*PI*i)^(k/n) = (cos(2*PI) + i*sin(2*PI))^(k/n) = (1+i*0)^(k/n) = 1
for all k,n

clearly there is something that i missing. i don't think its the algebra because the exponent rules are the same with the complex number as the
real numbers, but i can't figure out what is it.
hoping for your help and thank you very much !

 

[topsquark]

first of all i'm sorry about my english.
i'm learning computer science and now im studing about the FFT algorithm which using the comlex unitiy roots, and i came to a strange conclusion.
if
e^((2*PI*k*i)/n) = cos(2*PI*k/n) + i*sin(2*PI*k/n)

[for i imaginary number, and k,n natural numbers]

for all k,n (from the Euler's formula), than why

(e^2*PI*i)^(k/n) = (cos(2*PI) + i*sin(2*PI))^(k/n) = (1+i*0)^(k/n) = 1
for all k,n

clearly there is something that i missing. i don't think its the algebra because the exponent rules are the same with the complex number as the
real numbers, but i can't figure out what is it.
hoping for your help and thank you very much !

I think I can clear this up by a demonstration.

Let k = 1 and n = 2. Then:
\(\displaystyle e^{2 \pi i k/n} = e^{\pi i} = cos( \pi ) + i~sin(\pi) = -1\)

\(\displaystyle \left ( e^{2 \pi i} \right )^{k/n} = \left ( e^{2 \pi i} \right )^{1/2} = 1^{1/2}\)

The last expression should be -1 as well. And it can be, for \(\displaystyle \sqrt{1} = \pm 1\)

The varying values of the solution given by the first calculation can be any of the possible square roots of 1. The difference in the second equation is that we could do this:
\(\displaystyle \left ( e^{2 \pi i + 2m \pi i} \right )^{k/n}\) for integer m. This will generate all the roots of 1 given a specific value of m. Try it for k = 1, n = 2, and m = 0, 1.

-Dan

 

[rtk]

I think I can clear this up by a demonstration.

Let k = 1 and n = 2. Then:
\(\displaystyle e^{2 \pi i k/n} = e^{\pi i} = cos( \pi ) + i~sin(\pi) = -1\)

\(\displaystyle \left ( e^{2 \pi i} \right )^{k/n} = \left ( e^{2 \pi i} \right )^{1/2} = 1^{1/2}\)

The last expression should be -1 as well. And it can be, for \(\displaystyle \sqrt{1} = \pm 1\)

The varying values of the solution given by the first calculation can be any of the possible square roots of 1. The difference in the second equation is that we could do this:
\(\displaystyle \left ( e^{2 \pi i + 2m \pi i} \right )^{k/n}\) for integer m. This will generate all the roots of 1 given a specific value of m. Try it for k = 1, n = 2, and m = 0, 1.

-Dan

thank you very much ! you helped me a lot !

 

[rtk]

I think I can clear this up by a demonstration.

Let k = 1 and n = 2. Then:
\(\displaystyle e^{2 \pi i k/n} = e^{\pi i} = cos( \pi ) + i~sin(\pi) = -1\)

\(\displaystyle \left ( e^{2 \pi i} \right )^{k/n} = \left ( e^{2 \pi i} \right )^{1/2} = 1^{1/2}\)

The last expression should be -1 as well. And it can be, for \(\displaystyle \sqrt{1} = \pm 1\)

The varying values of the solution given by the first calculation can be any of the possible square roots of 1. The difference in the second equation is that we could do this:
\(\displaystyle \left ( e^{2 \pi i + 2m \pi i} \right )^{k/n}\) for integer m. This will generate all the roots of 1 given a specific value of m. Try it for k = 1, n = 2, and m = 0, 1.

-Dan

i also have another question : why does \(\displaystyle \left ( e^{2 \pi i + 2m \pi i} \right )^{k/n}\) has a cycle of 2pi ?

 

[pka]

i also have another question : why does \(\displaystyle \left ( e^{2 \pi i + 2m \pi i} \right )^{k/n}\) has a cycle of 2pi ?

Here is some ease of notation: \(\displaystyle e^{\bf{i}\theta}=\exp(\bf{i}\theta)\).
Now you need to realize that \(\displaystyle \exp(\bf{i}\theta)=\cos(\theta)+\bf{i}\sin(\theta)\)
Moreover, from elementary trigonometry: for all integers n \(\displaystyle \cos(\theta)=\cos(\theta+2n\pi)~\& ~\sin(\theta)= \sin (\theta+2n\pi)\)

Therefore it follows at once that \(\displaystyle e^{\bf{i}(\theta+2n\pi)}=\exp(\bf{i}(\theta+2n\pi))\)

 

[rtk]

Here is some ease of notation: \(\displaystyle e^{\bf{i}\theta}=\exp(\bf{i}\theta)\).
Now you need to realize that \(\displaystyle \exp(\bf{i}\theta)=\cos(\theta)+\bf{i}\sin(\theta)\)
Moreover, from elementary trigonometry: for all integers n \(\displaystyle \cos(\theta)=\cos(\theta+2n\pi)~\& ~\sin(\theta)= \sin (\theta+2n\pi)\)

Therefore it follows at once that \(\displaystyle e^{\bf{i}(\theta+2n\pi)}=\exp(\bf{i}(\theta+2n\pi))\)

thank you very much. much clearer to me right now.