For $0< \epsilon < 1$, suppose we have complex numbers $f_{\epsilon}$ such that each $|f_{\epsilon}| =1$ and $$ \lim_{\epsilon \to 0} \frac{f_\epsilon - 1}{\epsilon} := a $$
exists.
Prove that $a$ is a purely imaginary number.
This statement comes from the last paragraph in the proof of Theorem 13.38 from Rudin's functional analysis book (this is the proof of Stone's theorem for semigroups of normal operators). I have tried working out the proof just using basic complex analysis, but I am not able to conclude any more than $\text{Re} f_\epsilon \to 1$ and $\text{Im}f_\epsilon \to 0$.
I am not sure if I am interpreting Rudin's explanation correctly, perhaps there is more that needs to be assumed about the $f_\epsilon$ to achieve the conclusion above, but Rudin does not explicitly state this? Within the context of the proof, each $f_\epsilon$ is the Gelfand transform of the semigroup member $Q(\epsilon) = e^{\epsilon A}$ (A being the infinitesmial generator of the semigroup). And in this last stage of the proof it is assumed that each $Q(\epsilon)$ is unitary.
Hints or solutions are greatly appreciated.
Write $f_{\epsilon}=cos(\theta_{\epsilon})+i\sin(\theta_{\epsilon})$. Then
\begin{align*} \lim_{\epsilon\to0}\theta_{\epsilon}&=0,\\ \Im(a)&=\lim_{\epsilon\to0}\frac{\sin(\theta_{\epsilon})}{\epsilon},\\ \Re(a)&=\lim_{\epsilon\to0}\frac{cos(\theta_{\epsilon})-1}{\epsilon}=\lim_{\epsilon\to0}\frac{cos(\theta_{\epsilon})-1}{\theta_{\epsilon}}\frac{\theta_{\epsilon}}{\sin(\theta_{\epsilon})}\frac{\sin(\theta_{\epsilon})}{\epsilon}=0\times1\times\Im(a)=0. \end{align*}