# Complex-Valued Functions, limits, and conjugates

1. Sep 15, 2013

### mathsciguy

I have a question about complex valued functions, say f(z) where z=x+iy is a complex variable.
Can every such complex valued function be represented by:
f(z)=u(x,y)+iv(x,y)?

Also, is the limit of the conjugate such a function equal to the conjugate of the limit of the function?
Something like:
lim[conjugate[f]] (?)= conjugate[lim[f]]

2. Sep 15, 2013

### Jolb

Yes. You can always split a complex valued function into its real and imaginary parts. If you had f(z) as a function ℂ→ℂ, then you can define Re[f(z)]=u(z)=u(x,y) and Im[f(z)]=v(z)=v(x,y) for real-valued functions u and v.

You can even write the real parts and imaginary parts of a complex expression w in a way that is easier to manipulate: (I like to use the notation w* to denote the conjugate of w.)
Re[w] = (w+w*)/2
Im[w] = (w-w*)/2

Let's apply the formula that you just suggested. We have:
[f(z)]*=[u(x,y) + i v(x,y)]* = u(x,y) - i v(x,y)

Therefore,
lim [(f(z))*]=lim[u(x,y) - i v(x,y)] = lim [u(x,y)] - lim [i v(x,y)] = lim[u(x,y))] - i lim[v(x,y)] = (lim[u(x,y)] + i lim[v(x,y)])*

Last edited: Sep 15, 2013
3. Sep 15, 2013

### mathsciguy

Thanks, that's pretty straightforward huh.

Anyway, I wonder if you guys can answer a barely tangential question? Let's say I have a complex valued function f(x) of a real variable. If the limit f(x) as x-> infinity is zero, are the derivatives of f(x) as x-> infinity also zero?

4. Sep 15, 2013

### Jolb

Well, when you ask about the derivative of a complex-valued function, you usually want the derivative to be defined in a special way that makes it independent of the direction along which you take the derivative. (Unlike say taking the gradient of a real-valued function.) Functions of this sort are called analytic, and I do think that if an analytic function approaches 0 as z→∞, then its derivative must as well. [PS: on second thought, I'm not totally sure of this and I can't find a quick reference. Maybe somebody could clarify on this particular point.]

However there are (non-analytic) real valued functions which disobey that. Like for example look at the function f(x) = sin(x2)/x. Clearly it converges to 0 but what about its derivative? You can just stick a z instead of x and you've constructed complex function which disobeys your suggestion.

Last edited: Sep 15, 2013
5. Sep 15, 2013

### mathsciguy

Hm, well, I also thought so. But see, I was trying to figure out how to show that an operator like L = i(d^3/dx^3) is hermitian for functions f(x) in the interval -(infinity)< x < (infinity) where as f approaches infinity, it also approaches zero. I tried using the straightforward method where I repeatedly do integration by parts but then I'd end up with boundary terms containing the derivatives of f at infinity (all the boundary terms should be zero, and L is hermitian).

Edit: Turns out I can just argue that since for the given boundary conditions, L = -i(d/dx) is hermitian, then L^3 = i(d^3/dx^3) is also hermitian.

Is it because:
if <f|L(g)> = <(L(f))|g> then <f|L^3(g)> = <L^3(f)|g> ?

Last edited: Sep 15, 2013
6. Sep 15, 2013

### Jolb

Well, since it sounds like you're doing Quantum Mechanics, then in QM, when we deal with bound states (rather than scattering states), we typically do add the assumption that the wavefunction AND its first derivative approach zero at infinity. Here is the best reference I can find: http://www.colorado.edu/physics/phys3220/3220_fa97/notes/notes3/3220_notes3_1.html
This is one bit of mathematical trivia which often gets swept under the rug. There is a good discussion about this here:
http://mathoverflow.net/questions/1...aints-of-a-wave-function-in-quantum-mechanics

7. Sep 15, 2013

### Jolb

I'm not so sure about your proof in your last post. Let me write hermittivity like this: If L is hermitian then L=Lt, where Lt is the adjoint of L, defined as the conjugate transpose: Lt=(L*)T where T denotes the transpose.

So the way you would prove that L3 is hermitian, using the assumption that L is hermitian, would look something like this:

(L3)t|ψ>=Lt Lt Lt |ψ>
= L L L |ψ> = L3|ψ>

When you form matrix elements <φ|L|ψ>, you're usually not proving general things about the operators since you're projecting them onto a lower-dimensional subspace.

8. Sep 15, 2013

### mathsciguy

Thanks, I think that kind of proof is something I would think of if I was working with matrices.