MHB Understanding the Concept of Double Conjugate in Complex Functions

[ognik]

Hi, I am not sure what $ {f}^{*}({z}^{*}) $ means?
I think the '*'s cancel, ie $ {f}(z) = u(x,y) + iv(x,y), \: \therefore f({z}^{*}) = u(x,y) - iv(x,y), , \: \therefore {f}^{*}({z}^{*}) = u(x,y) + iv(x,y) $ ?
Thanks

 

[Euge]

Hi ognik,

If $z$ is represented by the point $(x,y) \in \Bbb R^2$, then $z^* = (x,-y)$. So given $f(z) = u(x,y) + iv(x,y)$, we have $f(z^*) = u(x,-y) + iv(x,-y)$ and hence $f^*(z^*) = u(x,-y)-iv(x,-y)$.

 

[ognik]

Crystal! Thanks Euge.