Where Does e^(2*pi*i) Originate?

• sirwalle
In summary, the textbook says that 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. Basic property of exponents says that e^{a+b} = e^a e^b, so e^(2*pi*i) = e^z*e^(2*pi*i) = e^z*e^(2*pi*i). However, I don't know where from they
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.

Basic property of exponents:

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

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

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..

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

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.

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.

sirwalle said:
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.

Mentallic said:
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

Borek said:
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

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

Hurkyl said:
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?

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)

1. What is e^(2*pi*i)?

e^(2*pi*i) is a mathematical expression that represents a complex number with a magnitude of 1 and an angle of 2*pi radians. It is also known as the imaginary unit and is commonly denoted as i.

2. Where does e^(2*pi*i) come from?

e^(2*pi*i) comes from the mathematical constant e, which is approximately equal to 2.71828, and the trigonometric function cosine. When e^(2*pi*i) is evaluated, it results in the value of 1, which is the same as the cosine of 2*pi radians.

3. What is the significance of e^(2*pi*i)?

e^(2*pi*i) has several applications in mathematics, physics, and engineering. It is used to represent periodic functions, such as waves and oscillations, and is also used in the study of complex numbers and their properties.

4. How is e^(2*pi*i) related to the concept of rotation?

Since e^(2*pi*i) has a magnitude of 1 and an angle of 2*pi radians, it can be represented as a point on the unit circle in the complex plane. This point can be interpreted as a rotation of 2*pi radians, which is equivalent to a full circle.

5. Can e^(2*pi*i) be simplified?

Yes, e^(2*pi*i) can be simplified to 1 using Euler's formula, which states that e^(i*x) = cos(x) + i*sin(x). When x = 2*pi, the cosine term becomes 1 and the sine term becomes 0, resulting in e^(2*pi*i) = 1.

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