Complex conjugates of functions

1. Aug 25, 2013

brmath

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?

2. Aug 25, 2013

Mandelbroth

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 $f(z)\bar{f}(\bar{z}^{-1})=u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})u(x,y)+v(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2})v(x,y)+i (u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) v(x,y)-v(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) u(x,y))$.

3. Aug 25, 2013

brmath

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*).