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

  • Thread starter Thread starter member 508213
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the misconception that taking the natural logarithm of both sides of the equation e^(i2π) = 1 leads to the conclusion that i2π = 0. It is established that the complex exponential function is not injective, meaning it does not have a unique inverse. This results in multiple x values corresponding to the same y value, as demonstrated by the example where y = 1 corresponds to both x = 0 and x = 2πi. The discussion emphasizes the importance of understanding the properties of complex functions when attempting to invert them.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with the exponential function and its inverse, the natural logarithm
  • Knowledge of injective functions and the concept of one-to-one correspondence
  • Basic algebraic manipulation, including handling square roots and exponents
NEXT STEPS
  • Study the properties of complex functions, specifically the complex exponential function
  • Learn about injective and non-injective functions in mathematical analysis
  • Explore the implications of multi-valued functions, particularly in complex analysis
  • Investigate the relationship between logarithmic and exponential functions in the context of complex numbers
USEFUL FOR

This discussion is beneficial for mathematics students, educators, and anyone interested in deepening their understanding of complex analysis and the properties of exponential functions.

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

i2pi=0

?
 
  • Like
Likes 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 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K