0
$\begingroup$

I have a function with a constant term upfront and was wondering how to obtain the complex conjugate. The function is say:

$$(g(z))^* = (\frac{-i\Gamma}{2\pi}\times f(z))^* $$

Would the conjugation be conducted 'distributively' such that:

$$ \ \ \ \ \ \ \ \ \ \ \ \ (g(z))^* = (\frac{-i\Gamma}{2\pi})^*\times f(z)^* $$ $$ \rightarrow(g(z))^* = \frac{i\Gamma}{2\pi}\times f(z)^* $$

also in general for a complex function if a complex function is contained within another how is the conjugate computed, like this?

$$g(z) = h(f(z)) $$ $$ \rightarrow g(z)^* = (h(f(z)))^* $$ $$ \rightarrow g(z)^* = h^*(f^*(z)) $$

Sorry for the dual question , but thank you all for your time!

$\endgroup$
1
$\begingroup$

Concerning your first question, you are right. This follow from the equality $\overline{z\times w}=\overline z\times\overline w$.

Now, if $f$ is a function from $\mathbb C$ into itself, let us define $\overline f(z)$ as $\overline{f(z)}$. Then $\overline{h\circ f}=\overline h\circ f$, which, in general, is not the same thing as $\overline h\circ\overline f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.