# When is the conjugate of an analytic complex function analytic also?

Given a function $f(x)$ that is analytic in some domain $D$ in the complex plane, if the conjugate is $$\overline{f(x)} = \frac{c^2}{f(x)}$$ then it follows that the conjugate is analytic. I read in a Churchill that this is true. Why does it follow that the conjugate must be analytic?

• This does not seem a sensible notion of "conjugate function". Typo? Misunderstanding? – paul garrett Sep 26 '14 at 0:09
• I think I meant to say conjugate rather than conjugate function, sorry. By conjugate, I mean $\overline{z} = u - iy$ for a $z = u + iy$ in the complex plane. – mathjacks Sep 26 '14 at 0:13
• The conjugate is not analytic. Eg the $\overline {z}$ is not analytic. Look at the Cauchy Riemann equations. – Rene Schipperus Sep 26 '14 at 0:14
• The expression written for "conjugate" is not literally the complex conjugate of the output of the function $x\to f(x)$. Something is confused here... – paul garrett Sep 26 '14 at 0:15
• I guess my confusion comes from this example in Churchill: i.imgur.com/xmNMCOC.png . Am I not using the terms correctly? – mathjacks Sep 26 '14 at 0:16

Let's approach it in the reverse order as Churchill. Suppose $f(z)$ is analytic within some domain $D$. Then, for any constant $c$, the function $g(z)=c^2/f(z)$ is analytic within the same domain as $f(z)$ (save at zeroes of $f(z)$, but we won't need to worry about that caveat.) So we have two functions which certainly satisfy the Cauchy-Riemann equations on $D$.
Note that nothing has been said about $\overline{f(z)}=\dfrac{|f(z)|^2}{f(z)}$. But suppose that the modulus of $f(z)$ is some nonzero constant within $D$ i.e. $|f(z)|=c\neq 0$ for some $c\in\mathbb{C}$, for all $z\in D$. Under that particular assumption, the $g(z)$ defined previously must be identical to $\overline{f(z)}$---and yet, by construction, it must also be analytic.
What this implies is that both $f(z)$ and its conjugate are analytic within $D$. But then the Cauchy-Riemann equations demand that $f(z)$ must be constant on this domain. (This result is simple to prove, and is presumably the Example 3 to which Churchill alludes.) Hence there are no non-constant functions which are both analytic and have constant modulus on a given domain.