$ \newcommand{\Exp}{\operatorname{Exp}} $ Let $ A $ be a unital Banach algebra. For $ a \in A $, consider $$ \Exp(A) \stackrel{\text{def}}{=} \{ e^{a_{1}} e^{a_{2}} \cdots e^{a_{n}} ~|~ n \in \mathbb{N} ~ \land ~ a_{1},a_{2},\ldots,a_{n} \in A \}. $$ I need to show the following:
(a) $ \Exp(A) $ is a subgroup of $ G(A) $, where $ G(A) $ is the set of invertible elements of $ A $.
(b) Prove that for every $ b \in \Exp(A) $, there exists a continuous function $ F: [0,1] \to G(A) $ such that $ F(0) = \mathbf{1}_{A} $ and $ F(1) = b $.
(c) Assume that $ c \in A $ and $ \sigma(c) \subseteq \{ \lambda \in \mathbb{C} ~|~ \text{Re}(\lambda) > 0 \} $. Using complex analysis, show that there exists a ‘logarithm of $ c $ in $ A $’, i.e., $ c = e^{d} $ for some $ d \in A $.
(d) Prove that $ \Exp(A) $ is an open subset of $ A $ using the following guidelines. Let $ b_{0} \in \Exp(A) $. Suppose that $ b \in G(A) $ satisfies $ \| b - b_{0} \| < \| b_{0}^{-1} \|^{-1} $ and consider $ c := b b_{0}^{-1} $. By estimating $ \| c - \mathbf{1}_{A} \| $, prove that Part (c) is applicable to $ c $, so $ b = c b_{0} \in \Exp(A) $.
I have thought about this question for a long time, but I could only figure out Part (a) and have no clue on how to do the rest. Please help. Thank you.
(a) As you have already figured out how to do this part, I shall skip it. :)
(b) Suppose that $ b = e^{a_{1}} \cdots e^{a_{n}} \in \text{Exp}(A) $. Define $ f: [0,1] \to \mathcal{G}(A) $ as follows: $$ \forall t \in [0,1]: \quad f(t) \stackrel{\text{def}}{=} e^{t a_{1}} \cdots e^{t a_{n}}. $$ Then
$ f $ is continuous,
$ f(0) = \mathbf{1}_{A} $ and
$ f(1) = b $.
(c) As $ \sigma(c) $ is a compact subset of $ \Omega := \{ \lambda \in \mathbb{C} ~|~ \text{Re}(\lambda) > 0 \} $, we can easily find a positively oriented rectifiable Jordan curve $ \gamma: [0,1] \to \Omega $ such that $ \sigma(c) $ lies in the interior region of $ \gamma $. As $ \ln: \Omega \to \mathbb{C} $ is a holomorphic function, we can use the Holomorphic Functional Calculus to define $$ \ln(c) \stackrel{\text{def}}{=} \frac{1}{2 \pi i} \int_{\gamma} \ln(\zeta) \cdot (\zeta \cdot \mathbf{1}_{A} - c)^{-1} ~ d{\zeta}. $$ Then $ \ln(c) $ has the property that $ e^{\ln(c)} = c $.
(d) Would you like to try this one on your own? Use the given hints to prove that $ \sigma(c) \subseteq \Omega $.