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E^(i * x) not welldefined. Why? 
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#1
Nov412, 06:18 PM

P: 27

Hi, Just saw this as a step in an example that demonstrates the differentiability of holomorphic function. But I can't for the life of me figure out why e^(2iθ) is illdefined.



#2
Nov412, 06:27 PM

P: 125




#3
Nov612, 03:50 PM

P: 350

It is well defined as a function of a real variable theta. But as a function on the plane, considering theta=theta(z), it has a problem at z=0. Does this answer your question? It is hard to tell without more context.
As a function of z, this function is the complex conjugate of (z/z)^2 


#4
Nov812, 07:23 AM

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E^(i * x) not welldefined. Why?
The standard definition of "function" for real numbers requires that "if x= y, then f(x)= f(y)" i.e. that f is "welldefined". For functions of complex variables, that is simply too restrictive. "Functions" that we would like to be able to use, such as [itex]e^x[/itex] would no longer be "functions". So we drop that requirement.



#5
Nov812, 08:13 AM

P: 136

Some functions are given the requirement of Principal Value to make some functions of a complex variables become actual functions.
I.e , let z=Re^(ix), and restrict x to be in (pi,pi]. This restriction works since e^(ix)=e^(ix+i*2n*pi) for all integers n. On the other hand, it makes functions behave less like we would wabt them to. That is, Log(xy)=/=Log(x)+Log(y) generally for principal value logarithm. On the other hand, log(xy)=log(x)+log(y). (correct me if I'm wrong as I am just typing from memory) Edit: x,y is a complex number for the logarithm examples. 


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