complex conjugates of functions


by brmath
Tags: complex-analysis
brmath
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#1
Aug25-13, 12:36 PM
P: 329
I am looking at a complex function f(z) and want to know something about f(z)f*(1/z*). We could assume for now that f(z) is analytic or at least meromorphic. Are there any identities involving this product? Is there any way to decompose the f*(1/z*) into a function of f?
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Mandelbroth
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#2
Aug25-13, 03:23 PM
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Quote Quote by brmath View Post
I am looking at a complex function f(z) and want to know something about f(z)f*(1/z*). We could assume for now that f(z) is analytic or at least meromorphic. Are there any identities involving this product? Is there any way to decompose the f*(1/z*) into a function of f?
Caveat: You mean "a complex function ##f##." ##f(z)## is the image (or "output") of ##z## under the function ##f##. This is a common abuse of notation.

Suppose ##f(x+iy)= u(x,y)+iv(x,y)##, where ##u## and ##v## are real-valued functions. Then, we have that ##\bar{f}(\frac{1}{x-iy})=\bar{f}(\frac{x}{x^2+y^2}+i\frac{y}{x^2+y^2})=u(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2})-iv(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})##.

Thus, we have ##f(z)\bar{f}(\bar{z}^{-1})=u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})u(x,y)+v(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2})v(x,y)+i (u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) v(x,y)-v(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) u(x,y))##.
brmath
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#3
Aug25-13, 06:58 PM
P: 329
Hi, I'll try not to write f(z) when I mean f. I already know that 1/z* = z/|z|. I had hoped there are some standard indentities or inequalities involving f(z)f*(1/z*).


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