I am reading through Theorem 5.5 in Hislop and Sigal, and I have the following confusion about the proof for item 2.
My confusion is, why can you immediately conclude that
\begin{equation}\label{key}
Ran(A-\lambda)^{\perp} = Ker(A^{*}-\lambda)?
\end{equation}
Surely, wouldn't it be
\begin{equation}
Ran(A-\lambda)^{\perp} = Ker(A^{*}-\overline{\lambda})?
\end{equation}
where $ \overline{\lambda} $ denotes the complex conjugate of $ \lambda $.
The result in Chapter $ 4 $ they refer to I believe is the standard result of $ \mathcal{H} $ being split into the orthogonal sum $ \overline{Ran A} \oplus Ker(A^{*}) $, and possibly the result that \begin{equation} \sigma(A^{*}) = \overline{\sigma(A)}, \end{equation} if $ A $ is closed. Maybe the result \begin{equation} Ran(A-\lambda)^{\perp} = Ker(A^{*}-\lambda) \end{equation} follows from \begin{equation} \sigma(A^{*}) = \overline{\sigma(A)}? \end{equation}
Thank you!
Edit: Sorry, I should have clarified that, it is certainly true if you accept item (1) in the theorem.
However, I am still confused because they say that item (1) will be proven in Theorem 5.6. 
.
In Theorem 5.6 they then refer back to equation 5.4, so it appears to me as if there is circular logic. What am I missing?
In (2), you want to prove that a subset of the spectrum (namely the residual spectrum) is empty, so it suffices to consider $\lambda$ in the spectrum, which is a subset of the reals. Of course then $\lambda=\bar \lambda$.
In any case, the identity $\operatorname{ran}(A-\lambda)^\perp=\ker(A-\lambda)$ also holds for $\lambda\in \mathbb{C}\setminus \mathbb{R}$ simply because the spaces on both sides are $\{0\}$ for every element not in the spectrum of $A$.