I am looking at the following post: Absolute Value of Complex Integral. I am wondering the following: if $f$ is continuous and $f(t) = r(t)e^{i\theta(t)}$ and we have strict equality:
\begin{equation} |\int_a^bf(t)dt| = \int_a^b|f(t)|dt \end{equation}
Is it true that$ |\int_a^b e^{i\theta(t)}dt| = \int_a^b|e^{i\theta(t)}|dt = b- a$?
This is true if $f(t) \neq 0$ for all $t$. For a counter-example note that if $f=0$ then $r(t)=0$ and $\theta (t)$ is arbitrary. Take $\theta (t)=t$, $a=0$ and $b =2\pi$ then $\int_a^{b} e^{i\theta (t)} d\theta =0 \neq b-a$.
Proof when $f(t) \neq 0$ for all $t$: Let $s=|\int_a^{b} f(t)dt|$. Then $\int_a^{b} f(t)dt=se^{ic}$ for some real number $c$. We get $\int_a^{b} r(t)dt =s$ from the hypothesis. Hence $\int_a^{b} e^{-ic}f(t)dt =\int_a^{b} r(t)dt$. Taking real parts we get $\int_a^{b} (r(t)-\Re (e^{-ic}f(t)))dt=0$. The integrand is non-negative and continuous. Hence $r(t)= r(t) \Re (e^{-ic}e^{i\theta (t)})$ for all $t$. Since $r(t) \neq 0$ this gives $ \Re (e^{-ic}e^{i\theta (t)})=1$ for all $t$ which implies $e^{i\theta(t)}=e^{ic}$ for all $t$. The conclusion is now obvious.