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