Where Does e^(2*pi*i) Come From?

[sirwalle]

Homework Statement

The problem at hand is that I don't understand wherefrom my textbook got a certain term(e^(2*pi*i). It doesn't say. At least not as I understand it.
The book says:

Homework Equations

e^(z+2*pi*i) = e^z*e^(2*pi*i) = e^z*1 = e^z

From where does e^(2*pi*i) come? I get the stuff leading to the answer, I just can't seem to understand from where that term comes from.

 

[Borek]

Basic property of exponents:

$$e^{a+b} = e^a e^b$$

 

[Vivamus]

I'm not certain as to what exactly your asking, but I hope this helps!

Recall the identity property of exponents:

e^a+b=e^ae^b

In your particular case, let:
a=z
b=2*pi*i

Therefore,
e^z+(2*pi*i) = e^z*e^2*pi*i

 

[sirwalle]

Oh, no, I am sorry if I was not clear. I simply don't know wherefrom they get the 2*pi*i from in e^(z+2*pi*i).

The information I get is what I've written. I believe that the 2*pi refers to the period. It just seems kind of abrupt to randomly insert it without any proof or reference to hardly anything..

 

[Bohrok]

From calculus one learns that $$e^{i\pi} = -1$$
So, using a certain property of exponents, $$e^{2i\pi} = (e^{i\pi})^2 = (-1)^2 = 1$$

 

[Borek]

OK. Could be they just added 2*pi*i at random. Why? Because they can

Do you know Euler's formula?

$$e^{ix} = \cos x + i \sin x$$

If you combine it all you see if you insert 2*pi*i into exponent at random, you will not change the result. Sometimes it can be a useful identity.

 

[Mentallic]

Actually they could add any integer multiple of $2\pi i$ and still leave the answer unchanged.

$$e^z=e^{z+2\pi i n}$$ where n is any integer.

 

[Hurkyl]

I just can't seem to understand from where that term comes from.

We can't directly help you if you don't show us what they were doing up to that point.

 

[Borek]

Actually they could add any integer multiple of $2\pi i$ and still leave the answer unchanged.

Lol, it must have been a senior moment on my side. I intended to write 2*pi*i*n but looked at 2*pi*i and decided there already is an integer (i) in the formula :grumpy:

 

[Mentallic]

Lol, it must have been a senior moment on my side. I intended to write 2*pi*i*n but looked at 2*pi*i and decided there already is an integer (i) in the formula :grumpy:

People seem to find new uses for i each and every day

 

[sirwalle]

We can't directly help you if you don't show us what they were doing up to that point.

That's the thing. They aren't doing anything, it has its own little "information box". It says nothing after, nothing before. Just what I've written. All I know is that it has to do with Euler (the chapter is about Euler), if that helps?

 

[Borek]

So it must be what I told you earlier - they just show an interesting and important property.

It is like asking where did the 2*pi came from in sin(x+n*2*pi) = sin(x) :tongue2: