I'm working through the following proof in C* algebras by Murphy, and I'm stuck on a step in the proof. For reference, $\epsilon_n = z^n : T \longrightarrow \mathbb{C}$, and $\Gamma = \text{span}(\epsilon_n)_{n \in \mathbb{Z}}$. Here is the current theorem I'm working through: 
Here is the theorem they are referencing:
I don't understand how they can conclude that $T_{e^{\psi''}}$ is invertible by the same argument for $T_{e^{\psi'}}$. We know that $e^{\psi'} \in H^{\infty}$, so by theorem $3.5.6$, this operator is clearly invertible, but we only know that $e^{\overline{\psi}''} \in H^{\infty}$, not $e^{\psi''}$, so we can only conclude that $e^{\psi''} \in L^{\infty}(T)$, and so the commutivity property claimed by theorem $3.5.6$ doesn't hold.
What mistake am I making here? Thank you.
I guess you work with the conjugate function $\overline{\psi''}$ and deduce that $T_{e^{\overline{\psi''}}}$ is invertible. Then you use that the adjoint of an invertible operator is invertible, and that $(T_f)^*=T_{\overline f}$.