Dirac delta function on the complex plane?

pellman

Supposedly,

&int; ez*(z - z0)f(z) dz*dz

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

(1/2&pi;)&int; eiy(x - x0)f(x) dxdy
= &int; &delta;(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

&int; ez*(z - z0)dz* = &delta;(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?

fresh_42

Mentor
2018 Award
$z^*$ 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 $f$.)

"Dirac delta function on the complex plane?"

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