# $\frac 1 {\overline {f(1/\ \overline z)}}$ is differentiable for $|z|>1$

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].
• 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 $g'(z) = \overline{f'(\bar z)}$. – Daniel Schepler Jan 24 '19 at 1:06
• Though actually, if $f$ is holomorphic in $D$ then $\overline{f(\bar z)}$ is holomorphic in $\bar D$. – Daniel Schepler Jan 24 '19 at 1:07