- #1

- 329

- 34

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter brmath
- Start date

- #1

- 329

- 34

- #2

- 611

- 24

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

- 329

- 34

Share: