# Homework Help: E^2*pi*i, where from?

1. Sep 19, 2010

### sirwalle

1. The problem statement, all variables and given/known data
The problem at hand is that I don't understand wherefrom my text book got a certain term(e^(2*pi*i). It doesn't say. At least not as I understand it.
The book says:

2. Relevant 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.

2. Sep 19, 2010

### Staff: Mentor

Basic property of exponents:

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

3. Sep 19, 2010

### Vivamus

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

Recall the identity property of exponents:

ea+b=eaeb

a=z
b=2*pi*i

Therefore,
ez+(2*pi*i) = ez*e2*pi*i

4. Sep 19, 2010

### 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..

5. Sep 19, 2010

### Bohrok

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

6. Sep 20, 2010

### Staff: Mentor

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.

7. Sep 20, 2010

### 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.

8. Sep 20, 2010

### Hurkyl

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

9. Sep 20, 2010

### Staff: Mentor

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:

10. Sep 20, 2010

### Mentallic

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

11. Sep 21, 2010

### sirwalle

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?

12. Sep 21, 2010

### Staff: Mentor

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: