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brmath

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- Thread starter brmath
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brmath

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- #2

Mandelbroth

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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))##.

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brmath

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