Do all complex functions have orthogonal real and imaginary parts?

[cocopops12]

z = h(x) + ig(x)
True or False: By the definition of the complex plane, h(x) and ig(x) will always be orthogonal.

If this was true, wouldn't that mean that we can find a 'very general' Fourier series representation of any function f(x) as an infinite series of An*h(x) + infinite series of Bn*ig(x) ?. I am aware that finding a Fourier series representation of f(x) doesn't mean that it will converge, and if it does converge, it won't necessary converge to f(x).

for example h(x) = x^2 , g(x) = ln(x)

Sorry if this is a stupid question, I'm just trying to understand some stuff...

Thanks.

 

[marcusl]

No, only analytic functions are orthogonal. Most functions are not analytic. Look at the real and imaginary parts of z'(z) = z^2, as one example.

 

[cocopops12]

I see, thank you sir!

 

[Bacle2]

No, only analytic functions are orthogonal. Most functions are not analytic. Look at the real and imaginary parts of z'(z) = z^2, as one example.

Maybe I misunderstand what you meant, but z^2 is analytic.

 

[marcusl]

Maybe I misunderstand what you meant, but z^2 is analytic.

Oops, sure enough that's my mistake!

 

[Bacle2]

No problem, happens to all of us.

 

[LayMuon]

No, only analytic functions are orthogonal. Most functions are not analytic. Look at the real and imaginary parts of z'(z) = z^2, as one example.

z^{1/2}

 

[Bacle2]

z^{1/2}

But z^1/2 has a region where it is analytic; it is not entire ( it is

actually a multi-function) , but you can find a region where it is analytic.

Use, e.g., the inverse function theorem to see that there are points for which

a local inverse exists. This local inverse is analytic in a 'hood of the point.

An example of a nowhere-analytic function is z^ , the conjugate function;

with z^(x+iy):= x-iy . Then z^:=U+iV , with U(x,y)=x and V(x,y)=-y

Let's use C-R:

U_x =1 , V_y=-1 , so U_x=V_y never holds.