# E^(-i * x) not well-defined. Why?

1. Nov 4, 2012

### nigels

e^(-i * x) not well-defined. Why??

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 ill-defined.

2. Nov 4, 2012

### Hetware

Re: e^(-i * x) not well-defined. Why??

What do you mean it is ill-defined. Why do you say that? It's well defined in the opinions of many smart people. I can't go beyond that until I know what objection there is to the conventional definition there is.

3. Nov 6, 2012

### Vargo

Re: e^(-i * x) not well-defined. Why??

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. Nov 8, 2012

### HallsofIvy

Re: e^(-i * x) not well-defined. Why??

The standard definition of "function" for real numbers requires that "if x= y, then f(x)= f(y)"- i.e. that f is "well-defined". For functions of complex variables, that is simply too restrictive. "Functions" that we would like to be able to use, such as $e^x$ would no longer be "functions". So we drop that requirement.

5. Nov 8, 2012

### Klungo

Re: e^(-i * x) not well-defined. Why??

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.