Theorem 1.9 If $\gamma$ is piecewise smooth and $f : [a,b] \to \Bbb{C}$ is continuous, then $$\int_{a}^{b} f d \gamma = \int_{a}^{b} f(t) \gamma '(t) dt$$
Here's part of the proof that confuses me:
Again we only consider the case where $\gamma$ is smooth. Also, by looking at the real and imaginary parts of $\gamma$, we reduce the proof to the case where $\gamma ([a,b]) \subseteq \Bbb{R}$
Why does it suffice to consider $\gamma$ real-valued? I can't quite figure this out. Presumably, we use the fact that $\gamma = Re ~ \gamma + i Im ~ \gamma$ and then use linearity of the Riemann-Stieltjes integral. But that seems like it would require that $Re ~ \gamma$ and $Im ~ \gamma$ be smooth functions, which in turns seems to require that the real and imaginary parts are differentiable, since it seems that we would need to use the chain-rule somewhere. But I don't believe that the real and imaginary parts are differentiable; they don't appear to satisfy the Cauchy-Riemann equations.
Again, this is all speculation on my part, since I don't quite understand the reduction step.
Yes, $\operatorname{Re}\gamma$ and $\operatorname{Im}\gamma$ are differentiable functions from $[a,b]$ into $\mathbb R$ (assuming that $\gamma$ is). This has nothing to do with the Cauchy-Rieamnn equations, since we are dealing with functions of a real variable here, not of a complex one.
If. for instance, $[a,b]=[0,2\pi]$ and $\gamma(t)=e^{it}$, then $\operatorname{Re}\gamma(t)=\cos t$ and $\operatorname{Im}\gamma(t)=\sin t$. So, $\operatorname{Re}\gamma$ and $\operatorname{Im}\gamma$ are differentiable functions, right?!