Fix $h \in (0,\pi/2)$. I am trying to explicitly exhibit a holomorphic function $f\colon \Sigma \to \mathbb{C}$, where $\Sigma$ is the punctured sector


$$\Sigma := \{z \in \mathbb{C} \:|\: z\neq 0, 0\leq\arg z\leq h\},$$


and where we require $f$ to satisfy the following conditions on $\partial\Sigma$:


* $f(x) \in \mathbb{R}$ for $x \in \mathbb{R}\_+$.
* $|f(e^{ih}x)|^2 = x^2 + 1$ for $x \in \mathbb{R}\_+$.


The only thing I tried so far is to start with $f(z) := z + 1$, which satisfies all the conditions except for having $|f(e^{ih}x)|^2 = x^2 + 2\cos(h) x + 1$, and to tweak that -- but I didn't get anywhere. I'd be happy for any tips, e.g. a proof that such an $f$ exists abstractly or doesn't exist at all. Best of all would be "$f$ exists, and here's a formula: ... ."


I did think twice about this being "research level", but, well, it is part of my research! I posted the same question at <https://math.stackexchange.com/questions/2766653/looking-for-holomorphic-function-on-a-sector-with-specified-boundary-behavior> , but didn't get any answers.



Looking for holomorphic function on a sector with specified boundary behavior
Such a function can be written with the help of the Poisson integral. 
First, your function must extend by reflection to $\{ z:|\arg z|\leq h\}$.
Let $g(z)=f((-iz)^{2h/\pi})$, then $g$ must be analytic in the upper half-plane, 
and 
$$g(t)=\sqrt{1+|t|^{4h/\pi}}.\quad\quad\quad\quad\quad (1)$$
To construct $g$,
first define a harmonic function $u$ using the Poisson integral:
$$u(x+iy)=\frac{y}{\pi} \int\_{-\infty}^\infty\frac{\log g(t)dt}{(t-x)^2+y^2},$$
where $g$ in the RHS is defined by (1), and let $v$ be a conjugate harmonic function, so that $g(z)=\exp(u+iv)$
is analytic in
the upper half-plane and has the desired value of $|g|=u$ on the real line. It remains to check that $g$ is real on the positive imaginary axis. This follows because 
$u(x+iy)=u(-x+iy),$ so we can choose $v$ so that $v$ is $0$ on the positive imaginary axis, therefore $g$ is positive on the positive imaginary axis.
This construction gives you $g$, and $f(z)=g(iz^{\pi/2h}).$


EDIT. The conjugate function can be also written as a formula. Combining it with Poisson's formula one can write 
$$\log g(z)=\frac{1}{\pi i}\int\_{-\infty}^\infty\frac{1+z^2}{1+t^2}\frac{\log g(t)dt}{t-z}+\frac{z}{\pi i}\int\_{-\infty}^\infty\frac{\log g(t)dt}{1+t^2}.$$
See, for example de Branges, [Hilbert spaces of entire functions](http://www.math.purdue.edu/~branges/Hilbert%20Spaces%20of%20Entire%20Functions.pdf), Thm. 2 on p. 3.


EDIT 2. Your function is not unique: one can multiply $g$ on any bounded function analytic in the upper half-plane, whose absolute value is $1$ on
the real line. Such functions can be completely described: they are of the form
$$e^{iaz}\prod\_{-\infty}^\infty\frac{z-a\_k}{z-\overline{a\_k}},\quad\quad\Im a\_k>0,\quad a\in{\mathbf{R}},
$$
where $(a\_k)$ is any sequence in the upper half-plane which satisfies the Blaschke condition (convergence of the product), and which is mapped by
$z\mapsto -\overline{z}$ onto itself.


Then the construction gives all such functions satisfying a mild growth restriction at $\infty$. 


