I am currently reading Hamming's Numerical Methods for Scientists and Engineers. On pg. 79 he discusses the topic of finding the zeros of a complex analytic function.

He then proceeds to discuss different types of complex conjugation for a function $w(z)$. Here are examples of the three types for $w(z)=\sin z = \frac{e^{iz} - e^{-iz}}{2i}$

  1. $\overline{w}(z)$ : replace $i$ with $-i$ in $w$, e.g. $\overline{w}(z)= \frac{e^{-iz} - e^{iz}}{-2i} = \sin z$.
  2. $w(\overline{z})$ : conjugate the argument, $w(\overline{z}) = \frac{e^{i\overline{z}} - e^{-i\overline{z}}}{2i}$
  3. $\overline{w(z)}$ : Hamming describes this as conjugating the values which I take to mean the conjugate of the image of $w(z)$ in the codomain. Although, I am not clear on this.

These three definitions have left me a bit confused. 2 and 3 seem relatively straight forward. But 1 leaves me a bit baffled in that it doesn't appear to be an actual complex conjugate of anything.

It appears that the first version can only be applied to functions of $z$, because if I rewrite

$w(z)=\sin z$ as $w(x + iy) = \sin(x+iy) = \sin x \cosh y + i \cos x \sinh y$


$\overline{w}(x + iy) = \sin x \cosh y - i \cos x \sinh y \neq w(x+iy) = \sin(x + iy)$

gives what I would expect to be the value of case 3 and a different answer then in the $z$-form where $w(z)=\overline{w}(z) = \sin z$.

Also, he later appears to state that $\overline{w(z)} = \overline{w}(\overline{z})$ although it is not clear whether that is true for all analytic functions or just those that have the property of $w(z) = \overline{w}(z)$ like $\sin z$.

My question is several-fold. Is the definition of conjugation given in 1 standard? What is its meaning? Also, is my interpretation of 3 correct and is $\overline{w(z)} = \overline{w}(\overline{z})$ the proper definition of 3.


The motivation behind understanding definition 1 is that Hamming uses it in the proof that analytic complex functions have zeros that are conjugate pairs if the function is real over the real domain.

He states without proof that if $w(z)$ is real for real $z$ then $w(z)=\overline{w}(z)$. He then provides the following proof for the above.


  • $\begingroup$ I don't understand all of what you write (there seems to be a mistake somewhere after "because if I rewrite", since $\sin(x+\mathrm iy)$ can't equal both of the right-hand sides), but regarding 1: If it's really meant to only replace all explicit occurrences of $\mathrm i$ by $-\mathrm i$ without replacing $z$ by $\bar z$, then yes, I'd agree that this isn't a form of conjugation; it's just some formal manipulation of the function with superficial similarity to conjugation, and denoting the result by $\overline w(z)$ seems potentially confusing. $\endgroup$
    – joriki
    Aug 26 '12 at 8:51
  • 1
    $\begingroup$ I fixed the mistake you mentioned. I intended to show that the type 1 conjugation only works in z-form because it gives a different result if the $w(z)$ is already written in the form $w(x+iy)=u(x,y) + i v(x,y)$. $\endgroup$
    – Jeff
    Aug 26 '12 at 14:27
  • 1
    $\begingroup$ Definition 1. is most natural if you think in series expansion around $z=0$: it conjugates the coefficients. $\endgroup$
    – WimC
    Aug 26 '12 at 14:48

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.

  • $\begingroup$ This answer looks interesting. It will take me some time to digest it but it looks like it gets to the heart of the matter. $\endgroup$
    – Jeff
    Aug 26 '12 at 20:45
  • 2
    $\begingroup$ @ChristianBlatter: This is exceptionally well explained (the entire explanation). I am wondering if you could provide a good source that explains this notion in a similar manner. $\endgroup$
    – TSGM
    Aug 3 '15 at 11:10
  • $\begingroup$ Which conjugation is it in the definition of an inner product on a function space? Can one use all of these? $\endgroup$
    – user123124
    Feb 9 '17 at 19:51

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 $\overline w(z)=\sum_{i=0}^\infty \overline{a_k} z^k$

  • $\begingroup$ I'll have to think about this definition to make sure it corresponds with Hamming's for cases other than $\sin z$ which is special in that $\overline{w}(z) = w(z)$ for it. $\endgroup$
    – Jeff
    Aug 26 '12 at 14:31
  • $\begingroup$ @Jeff: I hope my addition helps you with that. $\endgroup$
    – celtschk
    Aug 26 '12 at 16:59

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