Let $\Omega$ denote the open set $\{z\in \Bbb{C}: \Re(z)>0\}$. Let $(L_\lambda)_{\lambda\in\Omega }$ be a family of closed unbounded operators, with domain $D_{\lambda}\subset L^2(\Bbb{R}^2)$.
For $\lambda>0$ the spectrum of $L_\lambda$ is the set $\sigma_{\lambda}=\{(2m+1)\lambda, m\in \Bbb{N}\}$, so its resolvent $R_{\lambda}(\mu)=(L_{\lambda}-\mu I)^{-1}$ is defined for $\mu\not\in \sigma_{\lambda}$.
More precisely for $\lambda>0,\quad \mu\not\in \sigma_{\lambda},f\in L^2(\Bbb{R}^2) $; the resolvent $R_{\lambda}(\mu):L^2(\Bbb{R}^2)\to D_{\lambda} $ has the form $$R_{\lambda}(\mu)(f)(x,y)= \Gamma(\frac{\lambda-\mu}{2\lambda}) T_{\lambda}(\mu)(f)(x,y)$$
where $ T_{\lambda}(\mu): L^2(\Bbb{R}^2)\to D_{\lambda} $ is an bounded operator on $L^2(\Bbb{R}^2)$.
I have as assumptions:
1- For $\lambda\in\Omega$ and $\mu\in\Bbb{C}$ the operator $ T_{\lambda}(\mu): L^2(\Bbb{R}^2)\to D_{\lambda} $ is bounded on $L^2(\Bbb{R}^2)$.
2- For $\lambda\in\Omega$ and $f\in L^2(\Bbb{R}^2)$, the map is $\Bbb{C}\to \Bbb{C}:\mu\to T_{\lambda}(\mu)(f)$ holomorphic.
3- For $\mu\in\Bbb{C}$ and $f\in L^2(\Bbb{R}^2)$, the map $\Omega\to \Bbb{C}:\lambda\to T_{\lambda}(\mu)(f)$ is holomorphic.
First, let $H_1, H_2$ be Banach spaces. The set of bounded operators from $H_1$ to $H_2$ denoted by $B(H_1,H_2)$.
Definition A function $\Bbb{C}\supset U\ni z\to T_z\in B(H_1,H_2)$ is holomorphic if $\lim_{h\to 0}\frac{T_{z+h}-T_z}{h}$ exists for all $z\in U$.
In my case $H_1=H_2=H=\big(L^2(\Bbb{R}^2),\|\|_2\big)$ and $\|\|_2=\sqrt{\big< ., .\big>}$ where $\big< ., .\big>$ denotes the inner product of $H$.
Lemma0.1 Given operators $\Bbb{C}\supset D\ni z\to T_z, R_z \in B(H)= B(H,H)$ holomorphic on a domain $D$ (open and connected subset), if $T_z = R_z$ on some $V \subset D, V$ having an accumulation point, then $T_z = R_z$ on $D$.
Indeed, let $x,y\in H$. Define $f,g:D\to \Bbb{C}$ by $f(z)=\big< x, T_z(y)\big>$ and $g(z)=\big< x, R_z(y)\big>$. Hence $f$ and $g$ are holomorphic on $D [ \lim_{h\to 0}\frac{g(z+h)-g(z)}{h}= \big< x, \lim_{h\to 0}\frac{R_{z+h}-R_z}{h}(y)\big>$ exists for all $z\in D$ the same holds for $f$]. Since $f=g$ on $V$ then by the principle of the identity theorem $f=g$ on $D$. So $\big< x, T_z(y)\big>=\big< x, R_z(y)\big>$ for all $x\in H$ and so $ T_z(y)=R_z(y)$ for all $y\in H$. Thus $T_z=R_z$ on $D$.
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Apply this on my case. For this, note $S_{\lambda}(\mu)=(L_{\lambda}-\mu I)^{-1}; \lambda\in \Omega$ for suitable $\mu$.
Fix $\mu\in\Bbb{C}$.
By $3$ et $1$ of my assumptions and Lemma $0.2$, the map $S_{\lambda}(\mu)=(L_{\lambda}-\mu I)^{-1}: \Big[ \Omega\setminus\cup_{m\in\Bbb{N}}\big\{\frac{\mu}{2m+1}\big\}\Big]\longrightarrow B(H):\lambda \to S_{\lambda}(\mu)$ has the following forme $$S_{\lambda}(\mu)(f)(x,y)=\Gamma(\frac{\lambda-\mu}{2\lambda}) T_{\lambda}(\mu)(f)(x,y), \lambda\in \Big[ \Omega\setminus\cup_{m\in\Bbb{N}}\big\{\frac{\mu}{2m+1}\big\}\Big]$$.
Therefore, $\sigma_{\lambda}\subset\{(2m+1)\lambda, m\in\Bbb{N}\}$ for $\lambda\in\Omega$.
For the reverse inclusion. As an auxiliary result we have $\{(2m+1)\lambda, m\in\Bbb{N}\}\subset\sigma_{\lambda}$ for $\lambda\in\Omega$.
Consequently $$\sigma_{\lambda}=\{(2m+1)\lambda, m\in\Bbb{N}\}$$
My reasoning is it correct? i.e. Can we say that the spectrum of $L_{a+ib}$ is the set $\{(2m+1)(a+ib),m\in\Bbb{N}\}$?.
Thanks in advance.