Given a linear operator $A$ from a Hilbert space $\mathscr{H}$ to itself, we define the resolvent set as $$\rho (A) = \{\lambda \in \mathbb C\mid A-\lambda I\text{ is invertible}\}$$ I would like to show that the resolvent set of the adjoint is given by $$\rho(A^*)=\{\lambda\in\mathbb C\mid A^* -\lambda I\text{ is invertible}\}= \overline{\rho(A)} = \{\lambda\in\mathbb C\mid \bar{\lambda} \in \rho (A)\}$$ Here the bar denotes the complex conjugate. To show this, we have: $$\begin{align*} A^* -\lambda I\:\text{ is invertible} \\[1ex] & \iff (A^* -\lambda I)^*\:\text{ is invertible} \\[1ex] & \iff A - \bar{\lambda}I\:\text{ is invertible} \\[1ex] &\iff \bar{\lambda} \in \rho(A) \end{align*}$$
Hence $\lambda\in\bar\rho (A)\iff\lambda\in\rho (A^*)$.
Is this solution correct?
I would like to make sure, as the solution someone else gave me seems more complicated (rather needlessly) than this.