- #1
nonequilibrium
- 1,439
- 2
Hello.
I was wondering if you have a complex function [tex]f: A \subset C \to C[/tex] everywhere differentiable on A, can you say anything about the existence of [tex]\frac{d Re(f)}{d z}[/tex] or [tex]\frac{d Im(f)}{d z}[/tex]?
I was thinking that it's easy to proof that if [tex]f : A \subset R \to C[/tex], then Re(f) and Im(f) are differentiable, because
[tex]\frac{Re(f(x)) - Re(f(a))}{x-a} = Re \left( \frac{f(x) - f(a)}{x-a} \right)[/tex]
So [tex]\frac{d Re(f)}{d x} = Re \left( \frac{df}{dx} \right)[/tex]
Is that correct?
But generally, when [tex]A \subset C[/tex], I'm not sure how to find a counter-example or proof that it is always differentiable...
We do have:
[tex]\frac{f(x) - f(a)}{x-a} = \frac{Re(f(x)) - Re(f(a))}{x-a} + \frac{Im(f(x)) - Im(f(a))}{x-a}i[/tex], but all I can get out of that is that if either Re(f) or Im(f) is differentiable, then the other is also.
Any ideas?
EDIT: weird, I thought I had posted this in Analysis (which would seem more logical...), I hope this isn't a problem
I was wondering if you have a complex function [tex]f: A \subset C \to C[/tex] everywhere differentiable on A, can you say anything about the existence of [tex]\frac{d Re(f)}{d z}[/tex] or [tex]\frac{d Im(f)}{d z}[/tex]?
I was thinking that it's easy to proof that if [tex]f : A \subset R \to C[/tex], then Re(f) and Im(f) are differentiable, because
[tex]\frac{Re(f(x)) - Re(f(a))}{x-a} = Re \left( \frac{f(x) - f(a)}{x-a} \right)[/tex]
So [tex]\frac{d Re(f)}{d x} = Re \left( \frac{df}{dx} \right)[/tex]
Is that correct?
But generally, when [tex]A \subset C[/tex], I'm not sure how to find a counter-example or proof that it is always differentiable...
We do have:
[tex]\frac{f(x) - f(a)}{x-a} = \frac{Re(f(x)) - Re(f(a))}{x-a} + \frac{Im(f(x)) - Im(f(a))}{x-a}i[/tex], but all I can get out of that is that if either Re(f) or Im(f) is differentiable, then the other is also.
Any ideas?
EDIT: weird, I thought I had posted this in Analysis (which would seem more logical...), I hope this isn't a problem
Last edited: