Why Can't You Take the Ln of Both Sides of e^(i2(pi))=1 and Get i2pi=0?

  • Context: Undergrad 
  • Thread starter Thread starter member 508213
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the question of why taking the natural logarithm of both sides of the equation e^(i2(pi))=1 does not lead to the conclusion that i2pi=0. The scope includes mathematical reasoning related to complex numbers and the properties of functions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that taking the natural logarithm of both sides of e^(i2(pi))=1 leads to confusion because the complex exponential function is not injective, meaning it does not have a unique inverse.
  • Others illustrate the concept of non-injectivity by providing examples, such as the function f(x)=11, where different x values yield the same f(x), and the function y=x^2, where each nonzero y corresponds to two x values.
  • One participant mentions that while they initially thought they could use the natural log to keep the statements equivalent, they acknowledge a lack of deep understanding of complex numbers.
  • Another participant compares the situation to taking square roots, emphasizing that both the square root and logarithm functions are not single-valued, which leads to potential errors in reasoning.

Areas of Agreement / Disagreement

Participants generally agree on the non-injectivity of the complex exponential function and the implications for taking logarithms, but there is no consensus on the best way to articulate or understand these concepts.

Contextual Notes

Participants express varying levels of understanding regarding complex numbers and the properties of logarithmic and exponential functions, indicating that some assumptions about these functions may not be fully explored.

Who May Find This Useful

This discussion may be of interest to individuals studying complex analysis, mathematical functions, or those seeking clarification on the properties of logarithms and exponentials in the context of complex numbers.

member 508213
If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?
 
  • Like
Likes   Reactions: Mustapha
Physics news on Phys.org
$$e^{2\pi \, i}=e^{0}$$

we cannot conclude in general from

$$\mathrm{f}(x)=\mathrm{f}(y)$$
that
x=y

consider the function f(x)=11
f(67)=f(14.4)
but
67 is not equal to 11

(-3)^4=3^4
but
-3 is not equal to 3
 
Last edited:
lurflurf said:
$$e^{2\pi \, i}=e^{0\pi \, i}$$

we cannot conclude in general from

$$\mathrm{f}(x)=\mathrm{f}(y)$$
that
x=y

consider the function f(x)=11
f(67)=f(14.4)
but
67 is not equal to 11

(-3)^4=3^4
but
-3 is not equal to 3
what you say makes sense but I just thought that I would still be able to use the natural log and still keep the statements equivalent, but I do not understand complex numbers very deeply so I will just accept that what I did is not true.
 
Austin said:
If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?

When you take the ln of both sides, you're trying to invert the exponential function.

The problem is that the complex exponential function is not injective (its not one to one). In order for a function f\left(x\right)=y to have an inverse each x value must correspond to only one y value, and each y value must correspond to only one x value.

The complex exponential is not one-to-one. This is because there are y values that correspond to multiple x values. In your example y=1 corresponds to x=0 and x=2\pi i.

A similar thing happens with the function y=x^2. Again here each nonzero y value corresponds to two x values and a true inverse does not exist. You can take the square root to find the magnitude of x. But then you have to decide if you want the positive or negative root.
You can run into troubles if you take the the wrong root.

For instance consider the equation x^2 =4. One solution to this equation is x =-2. This means that I can write the equation as \left(-2\right)^2 =4. I could then take the square root of both sides giving me -2 =2. Obviously this is wrong. The problem is that I was very sloppy in how took the square root of (inverted) the function x^2. Its the same problem you run into when taking the ln of a complex exponential.
 
( to repeat wolfman): for the same reason you cannot take the square root of both sides of (1)^2 = (-1)^2 and get 1 = -1. i,e, sqrt is not a single valued function, and neither is ln. i.e. if g is not single valued, i.e. not a true function, you can have x = y but g(x) ≠ g(y).
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K