Modulus of Complex Integral

Theorem
Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \C$ be a continuous complex function.

Then:
 * $\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$

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:
 * $z = \ds \int_a^b \map f t \rd t$


 * $r \in \hointr 0 \to$ as the modulus of $z$


 * $\theta \in \hointr 0 {2 \pi}$ as the argument of $z$.

From Modulus and Argument of Complex Exponential:
 * $z = re^{i \theta}$

Then:

As $r$ is wholly real, we have:
 * $\ds 0 = \map \Im r = \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t$

Then:

As $\ds r = \size {\int_a^b \map f t \rd t}$ by its definition, the result follows.