(Complex Numbers) "Show that e^[(2n)(pi)(i)]=1 and e^[(2n+1)(pi)(i)]=-1, where n is an element of the set n_0."

I'm really not sure where to start on this, and I'm wondering if anybody who's completed similar problems would know what N_0 is. My book is incredibly poor at defining it, and I'm sure it's obviously related to the set of natural numbers, but I don't know if it's only odd, or even, or natural numbers and zero, etc.

Question

(Complex Numbers) "Show that e^[(2n)(pi)(i)]=1 and e^[(2n+1)(pi)(i)]=-1, where n is an element of the set n_0."

I'm really not sure where to start on this, and I'm wondering if anybody who's completed similar problems would know what N_0 is. My book is incredibly poor at defining it, and I'm sure it's obviously related to the set of natural numbers, but I don't know if it's only odd, or even, or natural numbers and zero, etc.

mendicant_bias · Answer

@SithsAndGiggles

anonymous · Answer

I think $N_0$ denotes the natural numbers, with the inclusion of $0$.

mendicant_bias · Answer

Alright, cool. But yeah, I don't really have a clue how to proceed here, could you give me a hint?

irishboy123 · Answer

just expand:ie  e^i theta = cos theta + i sin theta and it seems to fall into place.....

mendicant_bias · Answer

I'll try that; thanks. I'm not 100% sure what you say when you mean "expand", but I'll mess with Euler's formula.

rational · Answer

http://en.wikipedia.org/wiki/De_Moivre%27s_formula

anonymous · Answer

DeMoivre's theorem may also be of some help

mendicant_bias · Answer

$$e^{i(2 \pi n)}=\cos(2 \pi n)+i \sin(2 \pi n); \ \ \ \text{For all n, $i \sin(2 \pi n) = i(0)=0.$}$$$$e^{i(2 \pi n)}=\cos(2 \pi n)=1$$

Is this reasonable for the first part?

mendicant_bias · Answer

@SithsAndGiggles  (I'm not sure if people are getting notifications on posts from closed threads, not trying to spam you or anything)

anonymous · Answer

Yes, the proof for $2n+1$ will be nearly identical.

mendicant_bias · Answer

Yeah, I see now. Alright, awesome, thank you guys!