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As a school student I have seen a striking property of functions .

$$f\left(\bar{z}\right)=\overline{f\left(z\right)}$$

Where $z$ is a complex number and $\bar{z}$ it's complex conjugate. For eg: $z=x+iy$ then $\bar{z}=x-iy$ where $x,y \in \mathbb{R}$

How do we in generally prove the result ?

I have consulted some books on complex numbers and haven't yet seen a general proof, google search also didn't help, with this specific query. All places I have consulted just mentions this just like a physical law in nature (no questions regarding it's validity).But I believe only when we have a formal proof we could use it more power, and presently I don't believe that all functions obey this property .

If all functions do not obey the mentioned property(I have never came across a function that doesn't obey the result, but I agree that doesn't mean there doesn't exist one) , is there a specific name for such functions and how do we identify it without actually proving(I mean, proof not by showing that both the LHS and RHS are equal)

Proofs, references and any Google search tag's will be deeply appreciated

Thomas Andrews
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hrkrshnn
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    It's not true... take for example the constant function $f(z)=i$... – JP McCarthy Apr 04 '13 at 16:01
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    It's not true in general. For a simple exmaple, if you define $f(x+iy)=-y+xi=i(x+iy)$ it isn't true. – Thomas Andrews Apr 04 '13 at 16:01
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    If $f$ is analytic on some neighborhood of the real line and $f(z)$ is real whenever $z$ is real, you can show this is true, however. – Thomas Andrews Apr 04 '13 at 16:04
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    The application of this is usually limited to real polynomial functions $f(x)$, but it can be extended to a convergent power series with real coefficients. – hardmath Apr 04 '13 at 16:04
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    @ThomasAndrews: That additionally requires that the domain is (connected and) symmetric across the real line. Otherwise conjugating everything might cause different branches of the analytic continuation to be used at some points. – hmakholm left over Monica Apr 04 '13 at 16:08
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    Just to add a more pedagogical comment: conceptually it's often useful to think of functions as having almost no properties. By that I mean we can construct functions breaking essentially any conceivable identity of the kind in your question: we only know that $f(x)$ is a well-defined element of the range of $f$, but it can be anything at all in that range. Maybe this gives a bit of intuition as to why one might have been skeptical of the claim you made before actually constructing a counterexample. – Kevin Carlson Apr 04 '13 at 16:13

3 Answers3

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As mentioned in the comments, the result is not always true. But it is true for an interesting class of functions, namely those which can be written as a power series with real coefficients. Suppose $f(z) = \sum_{n=0}^{\infty} a_n z^n$ and $a_n$ are all real. Then

$$ \overline{f(z)} = \overline{ \sum_{n=0}^{\infty} a_n z^n}= \sum_{n=0}^{\infty} \overline{a_n z^n} = \sum_{n=0}^{\infty} \overline{a_n} \overline{z^n} =\sum_{n=0}^{\infty} a_n \overline{z}^n = f(\overline{z}). $$

To do those steps we used $\overline{x+y} = \overline{x} + \overline{y}, \overline{xy} = \overline{x}\overline{y}, \overline{z^n}=\overline{z}^n$ and that $\overline{x}=x$ if $x$ is real.

Included in this class are polynomials with real coefficients (and this fact is why polynomials with real coefficients have roots that come in complex conjugate pairs) and Sine/Cosine (whose power series have real coefficients).

Note to others: I've intentionally left out convergence issues, as I think it would hinder the OP more than it helps here.

Ragib Zaman
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If $f$ is analytic on some non-empty, connected, open set $D$ that is 'symmetric' in the sense that $\overline{D} = D$, then we have $f(\mathbb{R} \cap D) \subset \mathbb{R}$ iff $f(\overline{z}) = \overline{f(z)}$ for all $z \in D$.

First note that $D$ must intersect $\mathbb{R}$, since if $z \in D$, then $\overline{z} \in D$, and $D$ is path connected since it is open and connected.

($\Rightarrow$): Let $z_0 \in \mathbb{R} \cap D$. Then $f$ has a power series in some neighborhood of $z_0$, say $f(z) = \sum a_n (z-z_0)^n$. By restricting $z$ to the reals in the neighborhood, we have $f(z) \in \mathbb{R}$. We see (by differentiating and evaluating the result at $z_0$) that $a_n \in \mathbb{R}$ for all $n$, and so we have $f(\overline{z}) = \overline{f(z)}$ in the neighborhood of $z_0$.

A little work shows that the function $\phi(z) = f(z)-\overline{f(\overline{z})}$ is analytic on $D$, and the above shows that $\phi(z) = 0$ on an interval of the real line. Since $D$ is connected, it follows that $\phi(z) = 0$ on all of $D$.

($\Leftarrow$): If $z \in \mathbb{R}$, then $\overline{z} \in \mathbb{R}$, and so $f(z) = f( \overline{z} )$. It follows that $f(\mathbb{R} \cap D) \subset \mathbb{R}$.

copper.hat
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I don't think such function has any single word that describe them. If one were to speak about them without formulas I would call them "the functions that commute with complex conjugation".

Here's one example of an otherwise fairly nice function that doesn't satisfy this property: The function $f(z)=\frac{2z}{z^2+1}$ is holomorphic on the domain $$D=\mathbb C\setminus\bigl(\{x+i\mid x\ge 0\}\cup\{x-i\mid x\le 0\}\bigr).$$ Because $D$ is open and simply connected, $f$ has an antiderivative $F$ on $D$; by choosing an appropriate constant of integration we can select $F$ such that $F(0)=0$. Then $F(x)$ maps reals to reals -- in fact $F(x)=\log(x^2+1)$ for real $x$ -- and $F$ is holomorphic throughout its domain. But $F(a-bi)$ differs from $\overline{F(a+bi)}$ by $\pm 2\pi i$ whenever $|b|>1$.

hmakholm left over Monica
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