Complex Conjugate of Complex function

Solution 1:

In the complex world there is a fundamental involution, namely the map $\gamma:\ z\mapsto\bar z$. An involution of some set $X$ is a map $\iota:\ X\to X$ which is not the identity map ${\rm id}_X$, but whose square $\iota\circ\iota$ is the identity.

Given any region $\Omega\subset{\mathbb C}$ and a function $f:\ \Omega\to{\mathbb C}$, one can compose $f$ with $\gamma$ in various ways. Your cases 2. and 3. produce for a given $f$ the functions $$f\circ \gamma:\quad z\mapsto f\bigl(\bar z)$$ and $$\gamma\circ f:\quad z\mapsto\overline{f(z)}\ .$$ When $f$ is a holomorphic function of $z$ then both $f\circ\gamma$ and $\gamma\circ f$ are antiholomorphic, which means, e.g., that nonoriented angles between tangent vectors at points $z_0\in\Omega$ are preserved, but the orientation of small circles around $z_0$ is reversed.

Most interesting is your case 1. Here the function $f$ is transformed (also termed conjugated) into the new function $$\bar f:=\gamma\circ f\circ\gamma:\quad z\mapsto\overline{ f(\bar z)}\ .$$ When $f$ is a holomorphic function on $\Omega$ then it easy to check by means of the CR-equations and the chain rule that the function $\bar f$ is a holomorphic function on $\bar\Omega:=\{\bar z|z\in\Omega\}$.

The case when $\Omega$ and $\bar\Omega$ share an interval on the real axis is of special interest, because then we can ask the question: Could it be that in fact $\bar f(z)\equiv f(z)$ for all $z\in \Omega\cap\bar\Omega\ $? After all, in the definition of $\bar f$ there are two conjugations involved.

To investigate this case, assume that $0\in\Omega\cap\bar\Omega$ and that $$f(z)=\sum_{k=0}^\infty a_kz^k\qquad(|z|<\rho)$$ with certain complex coefficients $a_k$. Then $$\bar f(z)=\overline{\sum_{k=0}^\infty a_k(\bar z)^k}=\sum_{k=0}^\infty \bar a_k z^k\ ,$$ and this is $\ \equiv f(z)$ iff all $\bar a_k=a_k$, i.e., if all $a_k$ are in fact real. When the latter is the case then automatically $f(z)$ is real for real $z$, and it is not difficult to show the converse: If a holomorphic $f(z)$ is real for real $z$ (as in the case $f:=\sin$) then $\bar f=f$. This is the so-called reflection principle.

Solution 2:

Well, if you accept $w(\overline z)$ and $\overline{w(z)}$, you can simply define $\overline w(z)=\overline{w(\overline z)}$.

This allows also to resolve your example with $\sin z$: $$\begin{aligned} \overline\sin\, z &= \overline{\sin(\overline z)}\\ &= \overline{\sin(\overline{x+\mathrm iy})}\\ &= \overline{\sin(x-iy)}\\ &= \overline{\sin x\cosh y-\mathrm i\cos x\sinh y}\\ &= \sin x\cosh y+\mathrm i\cos x\sinh y\\ &= \sin(x+\mathrm iy)\\ &= \sin z \end{aligned}$$

It also directly follows that $\overline w(\overline z)=\overline{w(\overline {\overline z})}=\overline{w(z)}$.

Note that it may be easier to understand if we introduce the complex conjugation in function notation, $\mathrm{conj}(z) = \overline z$. Then $w(\bar z)=w(\mathrm{conj}(z)) = (w\circ\mathrm{conj})(z)$. Similarly, $\overline{w(z)} = (\mathrm{conj}\circ w)(z)$, and $\overline w(z)=(\mathrm{conj}\circ w\circ\mathrm{conj})(z)$.

Note that with the definition above, we have the following properties for $w(z)$:

• If $w(z)=f(z)+g(z)$, then $\overline w(z)=\overline f(z)+\overline g(z)$.

  Proof: $\overline w(z)=\overline{w(\overline z)} = \overline{f(\overline z)+g(\overline z)}=\overline{f(\overline z)}+\overline{g(\overline z)} = \overline f(z)+\overline g(z)$.

• If $w(z)=f(z)\cdot g(z)$, then $\overline w(z)=\overline f(z)\cdot\overline g(z)$.

  Proof: $\overline w(z)=\overline{w(\overline z)} = \overline{f(\overline z)\cdot g(\overline z)}=\overline{f(\overline z)}\cdot\overline{g(\overline z)} = \overline f(z)+\overline g(z)$

• If $w(z)=f(g(z))$, then $\overline w(z)=\overline f(\overline g(z))$.

  Proof: $\overline w(z) = \overline{w(\overline z)} = \overline{f(g(\overline z))} = \overline{f(\overline{\overline{g(\overline z)}})}=\overline{f(\overline{\overline g(z)})} = \overline f(\overline g(z))$

• If $w(z)=z$ then $\overline w(z)=z$.

  Proof: $\overline w(z) = \overline{w(\overline z)} = \overline{\overline z} = z$

• If $w(z)=\overline z$ then $\overline w(\overline z)=\overline z$.

  Proof: $\overline w(\overline z) = \overline{w(\overline{\overline z})} = \overline{w(z)} = \overline z$

• If $w(z)=c$ (where $c$ is a constant), then $\overline w(z)=\overline c$.

  Proof: $\overline w(z) = \overline{w(\overline z)} = \overline c$

Using these rules recursively, you especially get that if $w(z)=\sum_{i=0}^\infty a_k z^k$, then $\overline w(z)=\sum_{i=0}^\infty \overline{a_k} z^k$