Dirac delta function on the complex plane?

  • Thread starter pellman
  • Start date
  • #1
pellman
684
5
Supposedly,

∫ ez*(z - z0)f(z) dz*dz

is proportional to f(z0) much in the same way that

(1/2π)∫ eiy(x - x0)f(x) dxdy
= ∫ δ(x - x0)f(x) dx
= f(x0)


Is this true? Could someone help convince me of it, or point me to a text?

I would say that even if true, it would be incorrect to say that

∫ ez*(z - z0)dz* = δ(z - z0)

because the integration over dz and dz* cannot be done independently in the same way that a surface integral over dxdy in the plane can (sometimes) be separated into independent integrations over x and y. Or can it?
 

Answers and Replies

  • #2
fresh_42
Mentor
Insights Author
2021 Award
17,570
18,025
##z^*## is only a function in ##z##, too. So what you have is ##\exp(F(z))## with ##F(z)=G(z)(z-z_0)## where ##G(z)## is the conjugation.

(I used capital letters in order to avoid confusion with your function ##f##.)
 

Suggested for: Dirac delta function on the complex plane?

  • Last Post
Replies
5
Views
454
  • Last Post
Replies
2
Views
658
Replies
6
Views
168
Replies
2
Views
696
Replies
6
Views
966
Replies
2
Views
718
  • Last Post
Replies
3
Views
20K
Replies
3
Views
364
  • Last Post
Replies
24
Views
439
  • Last Post
Replies
2
Views
370
Top