Trouble in understanding why function is constant in a given domain

Definition The group of unitary operators $$u(\theta)$$ on $$L^2(\mathbb R^3)$$ given by $$(u(\theta)\Phi)(r) =\Phi(\theta) \equiv e^{\frac{3\theta}{2}}\Phi(e^\theta r)$$ is called the group of dilation operators on $$\mathbb R^3.$$

Let $$H$$ be a compact operator and $$R(z) \equiv(H-z)^{-1}$$ the resolvent. Define $$R(z,\theta)$$ by $$R(z,\theta) \equiv u(\theta)R(z)u(\theta)^{-1}$$. Let us define also $$\mathcal O \equiv \{\theta \in \mathbb C: (u(\theta)\Phi)(r)$$ for $$\theta \in \mathbb R$$ has an analytic continuation $$\}$$

In this article A Class of Analytic Perturbations for One-body Schrδdinger Hamiltonians they were able to show that the function $$\Psi_z(\theta)=(\phi(\theta) ,R(z,\theta)\phi(\theta))$$ is meromorphic in $$z$$ for $$z \in \mathcal C^{++} \equiv \{z \in \mathbb C : Im \ z > 0 ,\ Re \ z > 0 \}$$ and $$\theta \in \mathcal O^\epsilon \equiv \{\theta \in \mathcal O: Im \ \theta > \epsilon \}$$

Now since for $$\theta \in \mathbb R,\ u(\theta)$$ is unitary we have that

$$\Psi_z \equiv (\Phi,R(z)\Phi)=\Psi_z(\theta)$$

from this they claim that $$\Psi_z(\theta)$$ that for fixed $$z \in \mathcal C^{++}$$ and $$\theta \in \mathcal O^\epsilon$$ the function $$\Psi_z(\theta)$$ is constant in $$\theta$$.

My question is why is $$\Psi_z(\theta)$$ constant for fixed $$z \in \mathcal C^{++}$$ and $$\theta \in \mathcal O^\epsilon$$ ?

As I understand it, they have shown:

1. That $$\Phi_z(\theta)$$ is meromorphic in a region containing part of the real line.
2. That $$\Phi_z(\theta)$$ is constant on the real line (using unitarity of $$u(\theta)$$ for real $$\theta$$).

Hence by analytic continuation, $$\Phi_z(\theta)$$ must be constant over the entire region connected to the real line where it is meromorphic.

• So if a meromorphic function is constant in a interval than i will be constant in all its domain? Jan 12, 2020 at 3:07
• @amiltonmoreira In all of the domain which is connected to the interval, yes.
– Yly
Jan 12, 2020 at 3:08
• I gave a proof as question here math.stackexchange.com/questions/3505785/… Jan 12, 2020 at 3:09
• @amiltonmoreira Yes, your proof looks basically correct. It all follows from uniqueness of analytic continuation.
– Yly
Jan 12, 2020 at 3:13
• @amiltonmoreira Yes. The wikipedia page on meromorphic functions says so explicitly in the second paragraph: en.wikipedia.org/wiki/Meromorphic_function
– Yly
Jan 12, 2020 at 3:18