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

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In summary, taking the ln of both sides in the equation e^(i2(pi))=1 and having i2pi=0 is not valid because the complex exponential function is not one-to-one, meaning that one y value can correspond to multiple x values. This is similar to the function y=x^2 where one y value can correspond to two x values. Inverting the complex exponential function can lead to incorrect solutions, just like taking the square root of both sides in an equation like (1)^2 = (-1)^2.
  • #1
member 508213
If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?
 
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  • #2
$$e^{2\pi \, i}=e^{0}$$

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
 
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  • #3
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.
 
  • #4
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 [itex]f\left(x\right)=y[/itex] 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 [itex]y=1[/itex] corresponds to [itex]x=0[/itex] and [itex]x=2\pi i[/itex].

A similar thing happens with the function [itex]y=x^2 [/itex]. 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 [itex]x^2 =4 [/itex]. One solution to this equation is [itex]x =-2 [/itex]. This means that I can write the equation as [itex]\left(-2\right)^2 =4 [/itex]. I could then take the square root of both sides giving me [itex]-2 =2 [/itex]. Obviously this is wrong. The problem is that I was very sloppy in how took the square root of (inverted) the function [itex]x^2[/itex]. Its the same problem you run into when taking the ln of a complex exponential.
 
  • #5
( 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).
 

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

What is the meaning of "Solving E^i2pi = 1"?

The equation E^i2pi = 1 is known as Euler's identity. It is a mathematical equation that relates the values of the mathematical constants e, i, and pi. It is often considered one of the most beautiful and profound equations in mathematics.

Why is Euler's identity important?

Euler's identity is important because it connects three fundamental mathematical constants and shows their relationship in a single equation. It is also used in many areas of mathematics, physics, and engineering, and has implications in complex analysis and Fourier analysis.

How do you solve E^i2pi = 1?

To solve E^i2pi = 1, we can use the basic definition of exponential functions, where E^x is equal to e to the power of x. In this case, we can rewrite the equation as e^(i2pi) = 1. Since e^0 = 1, we can conclude that i2pi = 0. This means that i = 0, and therefore, the equation is satisfied for any value of i.

What does the solution of E^i2pi = 1 tell us about complex numbers?

The solution of E^i2pi = 1 tells us that complex numbers, represented by i, have a periodic nature. This means that when we raise e to any multiple of i2pi, we will always get the value of 1. This is a fundamental property of complex numbers and has many applications in mathematics and physics.

Can Euler's identity be used to prove other mathematical equations?

Yes, Euler's identity can be used to prove many other mathematical equations, such as De Moivre's formula and the relationship between sine and cosine functions. It also has applications in solving differential equations and understanding the behavior of periodic functions.

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