Let $E$ be a complex Hilbert space with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$.
If $A\in\mathcal{L}(E)$, why \begin{align*} \rho(A):=\mathbb{C}\setminus\sigma(A) & = \{\lambda \in \mathbb{C}\colon \exists d>0; \|(\lambda \mathrm{Id}-A)y\|\geqslant d\|y\|,\;\forall\;y\in E\\ &\phantom{+++}\;\hbox{and}\;\;\mathrm{dist}(x, \mathrm{Im}(\lambda\mathrm{Id}-A))=0\,\forall x\in E\;\}? \end{align*} where $\mathrm{dist}$ is the distance induced by $\|\cdot\|$.
The second condition says the image of $\lambda I - A$ is dense. The first implies $\lambda I - A$ is injective and has closed image. So $\lambda I - A$ is a linear bijection. The first condition also implies its inverse is bounded.