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I'm doing a bit of self study, but I'm uncomfortable with a certain idea. I want to show that $f(z)$ is analytic if and only if $\overline{f(\bar{z})}$ is analytic, and by analytic I mean differentiable at each point. Here $f$ is a complex valued function.

What I do is write $f(z)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real functions of two variables. Then $\overline{f(\bar{z})}=u(x,-y)-iv(x,-y)$.

These two forms look very similar, in the sense the one function being differentiable should immediately imply the other is differentiable, since the only thing really changing might be a $-$ sign popping out due to the chain rule.

How can I more rigorously express that $f(z)$ is analytic iff $\overline{f(\bar{z})}$ using this? Many thanks.

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4  
Analytic is equivalent to Cauchy Riemann equations + C^1. So verify the Cauchy-Riemann equations using a simple application of chain rule. –  Zarrax Jan 27 '12 at 5:23
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How would you show this nonrigurously? –  Mariano Suárez-Alvarez Dec 6 '12 at 23:35
    
Problem given by Ahlfors in Complex Analysis Page 28 –  Dutta Dec 28 '13 at 15:44

5 Answers 5

up vote 21 down vote accepted

Define $g(z)=\overline{f(\overline z)}$, where $f$ is defined on a domain that is symmetric about the real axis. Then note that $f(z)=\overline{g(\overline z)}$, so there is symmetry allowing only one implication to be shown directly. Suppose that $f$ is analytic. Then for all $z$,

$$\begin{align*}g'(z)&=\lim\limits_{h\to 0}\frac{g(z+h)-g(z)}{h}\\ &=\lim\limits_{h\to 0}\frac{\overline{f(\overline{z+h})}-\overline{f(\overline z)}}{h}\\ &=\lim\limits_{h\to 0}\overline{\left( \frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right) }\\ &=\overline{\left( \lim\limits_{h\to 0}\frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right) }\\ &=\overline{f'(\overline z)}. \end{align*}$$

That is, $g$ is differentiable, with $g'(z)=\overline{f'(\overline z)}$.


Alternatively, as other answers have indicated, you could check that the Cauchy-Riemann equations hold for $g$ if they hold for $f$, with $f(x+iy)=u(x,y)+iv(x,y)$ and $g(x+iy)=u(x,-y)+i(-v(x,-y))$ as you indicated.


Another perhaps more conceptual way to think of this is that complex analytic maps are conformal (where their derivatives are nonzero), preserving orientation and angles. Complex conjugation preserves angles but reverses orientation. Reversing orientation twice gets you back where you started, so the result is that $g$ is conformal. (I have given an idea here rather than anything close to a rigorous proof.)


Another approach is to look at power series expansions. If $f$ has power series expansion in a neighborhood of $\overline{c}$, $\displaystyle{f(z)=\sum\limits_{k=0}^\infty a_k(z-\overline c)^k}$, then in a neighborhood of $c$, $g$ has the power series expansion $\displaystyle{g(z)=\sum\limits_{k=0}^\infty\overline{a_k}(z-c)^k}$. That is, you just conjugate the coefficients and conjugate the base point for the expansion. This shows that $g$ is analytic if $f$ is.

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2  
Thanks Jonas, I was thinking about doing it the first way you show, but doesn't it require some justification to move the limit inside the complex conjugation? I wasn't sure how to provide it, so I shied away. –  Dedede Jan 27 '12 at 5:45
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@Dedede: Moving a limit inside conjugation is justified by continuity, and in fact conjugation preserves distances. Suppose that $\lim\limits_{h\to 0}F(h)=a$. If you want a rigorous definition, this means that for all $\varepsilon>0$ there exists $\delta>0$ such that $0<|h|<\delta$ implies $|F(h)-a|<\varepsilon$. But then $0<|h|<\delta$ implies $|\overline{F(h)}-\overline{a}|=|F(h)-a|<\varepsilon$, from which it follows that $\lim\limits_{h\to 0} \overline{F(h)}=\overline a$. This only uses the fact that $|\overline z|=|z|$ for all $z$. –  Jonas Meyer Jan 27 '12 at 5:52
    
@Dedede: By the way, this is the same reason conjugating a power series term-by-term is justified. (Also, where I wrote "only uses" in my last comment wasn't quite right, because it also uses $\overline{z-w}=\overline z-\overline{w}$ for all $w$ and $z$.) –  Jonas Meyer Jan 27 '12 at 6:15
    
Thanks Jonas, this was very comprehensive. –  Dedede Jan 27 '12 at 18:58

Hint: Cauchy-Riemann equations.

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Use the Cauchy-Riemann equations. Say $\overline{f(\bar{z})}=\alpha(x,y)+i\beta(x,y)$, so that you have

$$\alpha(x,y)=u(x,-y);$$ $$\beta(x,y)=-v(x,-y).$$

Thus

$$\frac{\partial\alpha}{\partial x}=u_x(x,-y),\qquad \frac{\partial\beta}{\partial y}=(-1)\cdot(-v_y(x,-y))=v_y(x,-y);$$

$$\frac{\partial\alpha}{\partial y}=(-1)\cdot u_y(x,-y),\qquad\frac{\partial\beta}{\partial x}= -v_x(x,-y).$$

Show that if one or the other of $\alpha,\beta$ and $u,v$ satisfies CR, so does the other pair.

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Thanks anon. So assuming it holds for $u$ and $v$, I can say: $$\frac{\partial\alpha}{\partial x}=u_x(x,-y)=v_y(x,-y)= \frac{\partial\beta}{\partial y}$$ and $$ \frac{\partial\alpha}{\partial y}=-u_y(x,-y)=v_x(x,-y)= -\frac{\partial\beta}{\partial x}$$? And the converse essentially follows from the same equalities? –  Dedede Jan 27 '12 at 5:42
    
@Dedede: Almost; $\displaystyle \frac{\partial\beta}{\partial x}=\color{Red}- v_x(x,-y)$. After that fix the converse follows from the equalities. –  anon Jan 27 '12 at 5:49

Let $K$ be a Hausdorff topological field, let $f$ be a $K$-valued function defined on an open subset $U$ of $K$, and let $a$ be a point of $U$.

Say that $f$ is differentiable at $a$ if there is a function $g$ from $U$ to $K$ which is continuous at $a$ and satisfies $$ f(z)=f(a)+(z-a)\ g(z) $$ for all $z$ in $U$. In this case, we write $f'(a):=g(a)$. (One easily checks that this makes sense.)

Assume that this condition holds, and that $\phi$ is an automorphism of $K$ viewed as a topological field.

Then $z\mapsto\phi(f(\phi^{-1}(z))$ is differentiable at $\phi(a)$, and we have $$ (\phi\circ f\circ\phi^{-1})'(\phi(a))=\phi(f'(a)). $$ The proof is straightforward.

EDIT. Here is the proof. By assumption we have $$ f(z)=f(a)+(z-a)\ g(z)\ \ \forall\ z\in U,\quad f'(a)=g(a). $$ Put $$ \widetilde f:=\phi\circ f\circ\phi^{-1},\quad \widetilde g:=\phi\circ g\circ\phi^{-1},\quad \widetilde a:=\phi(a),\quad \widetilde z:=\phi(z),\quad \widetilde U:=\phi(U). $$ This yields $$ \widetilde f(\widetilde z\,)=\widetilde f(\widetilde a\,)+(\widetilde z-\widetilde a\,)\ \widetilde g(\widetilde z\,)\ \ \forall\ \widetilde z\in\widetilde U,\quad\left(\widetilde f\ \right)'(\widetilde a\,)=\widetilde g(\widetilde a\,)=\phi(f'(a)). $$ This fashion of stating the definition of differentiability is due Carathéodory. See this answer.

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Another way: Morera's Theorem. Of course you have to figure our how two convert between integral $dz$ and integral $d\overline{z}$.

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