Triangle Inequality for Integrals/Complex

Theorem
Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f : \left[{a \,.\,.\, b}\right] \to \C$ be a continuous complex function.

Then:
 * $\displaystyle \left|{\int_a^b f \left({t}\right) \rd t}\right| \le \int_a^b \left|{f \left({t}\right)}\right| \rd t$

where the first integral is a complex Riemann integral, and the second integral is a definite real integral.

Proof
Define $z \in \C$ as the value of the complex Riemann integral:
 * $z = \displaystyle \int_a^b f \left({t}\right) \rd t$

Define $r \in \left[{0 \,.\,.\, \infty }\right)$ as the modulus of $z$, and $\theta \in \left[{0 \,.\,.\, 2 \pi}\right)$ as the argument of $z$.

From Modulus and Argument of Complex Exponential, we have $z = re^{i \theta}$.

Then:

As $r$ is wholly real, we have:
 * $\displaystyle 0 = \operatorname{Im}\left({r}\right) = \int_a^b \operatorname{Im} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t$

Then:

As $\displaystyle r = \left|{\int_a^b f \left({t}\right) \rd t}\right|$ by its definition, the result follows.