Let $S=\{z \in \mathbb C : |z|<1\}$. Suppose that $f: S \to \mathbb C$ is differentiable everywhere in its domain. Suppose further than $f(z)$ is never zero.

I want to show that $g(z):=\frac 1 {\overline {f(1/\ \overline z)}}$ is differentiable for $|z|>1$.

My first thought was to apply the composition-of-differentiable-functions-is-differentiable theorem. But, alas, the complex conjugation function is not differentiable.

I also thought of applying a theorem which states that, if $f$ is analytic in some open connected set $D$ which contains a segment of the x-axis and is symmertic about that axis, then $[\forall z \in D,\overline {f(z)}=f(\overline z)] \iff [f(x)$ is real for each $x$ in the segment$]$.

The above theorem would be helpful if $f(x)$ was real for each $x\in (-1,0) \cup (0,1)$, but this isn't necessarily true. For example, it isn't true if $f(z) \equiv iz$.


This is immediate from the fact that if $g$ is holomorphic in a domain $D$ the so is $\overline {f(\overline {z})}$ on it domain $\{\overline {z}:z \in D\}$. [one way to see this is to use power series. Another way is to verify C-R equations].

  • $\begingroup$ Or you could just go directly to the definition: if $g(z) = \overline{f(\bar{z})}$ then $\frac{g(z+h)-g(z)}{h}$ is the complex conjugate of $\frac{f(\bar z + \bar h) - f(\bar z)}{\bar h}$; but as $h \to 0$, also $\bar h \to 0$, so this quotient goes to $f'(\bar z)$ implying that $g$ is differentiable and $g'(z) = \overline{f'(\bar z)}$. $\endgroup$ – Daniel Schepler Jan 24 '19 at 1:06
  • $\begingroup$ Though actually, if $f$ is holomorphic in $D$ then $\overline{f(\bar z)}$ is holomorphic in $\bar D$. $\endgroup$ – Daniel Schepler Jan 24 '19 at 1:07
  • $\begingroup$ @DanielSchepler Thanks for the comment. $\endgroup$ – Kavi Rama Murthy Jan 24 '19 at 5:07

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