Prove e^(i2pin) = 1

prove e^(i*2pi*n) = 1
where n is just an integer

Using e^(ix) = cos(x) + isin(x)

I get:
e^(i*2pi*n) = cos(2pi*n) + isin(2pi*n)
= cos(2pi)^n + isin(2pi)^n
Im not sure what to do next...

I know the trig identity:
cos^2(x) + sin^2(x) = 1

Is this some way linked to proving it equals 1? and if so where does the i dissappear to?

(edited 11 years ago)

Reply 1

dknt

12

Original post by Sifr

prove e^(i*2pi*n) = 1
where n is just an integer

Using e^(ix) = cos(x) + isin(x)

I get:
e^(i*2pi*n) = cos(2pi*n) + isin(2pi*n)
= cos(2pi)^n + isin(2pi)^n
Im not sure what to do next...

I know the trig identity:
cos^2(x) + sin^2(x) = 1

Is this some way linked to proving it equals 1? and if so where does the i dissappear to?

By de Moivre's theorem it's (cos(2pi) + isin(2pi))^n, which I hope you can see is not necessarily the same as what you've written (you'll have various nth order cross terms, unless they evaluate to 0). The required result then follows, from considering what happens to cos and sin at multiples of pi.

As a note, by coincidence what you have written does give you the required result, but only because the cross terms evaluate to 0, which wouldn't necessarily be the case for some general angle. Also you can actually get the result without using de Movie's theorem, by using the periodic properties of cos and sin.

(edited 11 years ago)

Reply 2

electriic_ink

15

cos(2pi*n) = 1
sin(2pi*n) = 0

for any natural n.

Reply 3

davros

16

Original post by Sifr

prove e^(i*2pi*n) = 1
where n is just an integer

Using e^(ix) = cos(x) + isin(x)

I get:
e^(i*2pi*n) = cos(2pi*n) + isin(2pi*n)
= cos(2pi)^n + isin(2pi)^n

you've got a couple of nasty misconceptions in there!

cos(2\pi n) \neq cos(2\pi)^n

sin(2\pi n) \neq sin(2\pi)^n

As electriic_ink points out, you should be able to write down straightaway what

cos(2\pi n)

and

sin(2\pi n)

are from the periodicity of sin and cos.

Reply 4

ζ(x)

1

e^{2n i \pi} = \left(e^{i \pi}\right)^{2n} = (-1)^{2n} = 1.

Reply 5

Indeterminate

3

^^ As long as you know how Euler's formula (famously) simplifies in the case

x=\pi

it's a one-line proof for all integers n.

Reply 6

SaraWarah

Euler's.

ae^{ix} = z

where z is a complex number with modulus a and argument x.

Hence, in the case of

e^{in2\pi}

then on an Argand diagram (y-axis is "i", and the x-axis is just "x") what you have is a line at an angle of

n2\pi

from the positive x-axis.

All integer multiples of 2Pi are going to still be the same as 0 degrees, hence the complex number which results is just a line out that way -> with length 1.

On an Argand diagram, then, it's just 1.

(edited 11 years ago)