member 508213
If e^(i2(pi))=1 then why can't you take the ln of both sides and have
i2pi=0
?
i2pi=0
?
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.
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.
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.
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.
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.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
Austin said:If e^(i2(pi))=1 then why can't you take the ln of both sides and have
i2pi=0
?