I am trying to learn complex analysis using Alfohrs textbook and I have doubts about the proof of this property:
When $a\leq b$, the fundamental inequality
(1) $|\int_a^b f(t)dt| \leq \int_a^b |f(t)|dt$ holds for arbitrary complex $f(t)$.I will copy the proof given by Alfohrs:
To see this, we choose $c=e^{-i\theta}$ with a real $\theta$ and find $Re[e^{-i\theta}\int_a^b f(t)dt]=\int_a^b Re[e^{-i\theta}f(t)]dt\leq \int_a^b |f(t)|dt$. For $\theta=arg\int_a^bf(t)dt$ the expression on the left reduces to the absolute value of the integral, and (1) results.
I would appreciate if someone could clearly explain me or prove to me these two things: $\int_a^b Re[e^{-i\theta}f(t)]dt\leq \int_a^b |f(t)|dt$ I don't see and couldn't show this inequality by myself.
The second thing I couldn't grasp is: $\theta=arg\int_a^bf(t)dt \implies Re[e^{-i\theta}\int_a^b f(t)dt]=|\int_a^b f(t)dt|$
Maybe these two things are trivial but I really can't understand them, any explanations and proofs are very welcome.
We have $\int_a^b\operatorname{Re}(e^{-i\theta}f(t))dt \leq \int_a^b|f(t)|dt$ because
$$\operatorname{Re}(e^{-i\theta}f(t)) \leq \sqrt{\operatorname{Re}(e^{-i\theta}f(t))^2} \leq \sqrt{\operatorname{Re}(e^{-i\theta}f(t))^2 + \operatorname{Im}(e^{-i\theta}f(t))^2} = |e^{-i\theta}f(t)| = |f(t)|.$$
As for your second question, note that $\int_a^bf(t)dt \in \mathbb{C}$, so $\int_a^bf(t)dt = re^{i\theta}$ for some $r \geq 0$ and $\theta = \arg\int_a^bf(t)dt$. Then we have
$$e^{-i\theta}\int_a^bf(t)dt = e^{-i\theta}re^{i\theta} = r = \left|re^{i\theta}\right| = \left|\int_a^bf(t)dt\right|.$$
In particular, $\operatorname{Re}\left(e^{-i\theta}\int_a^bf(t)dt\right) = \left|\int_a^bf(t)dt\right|$.