- Proving the first inequality
Suppose, WLOG, that
$\int_{a}^{b}f(x)dx \neq 0$
Define
$ \lambda = \frac{\left| \int_{a}^{b}f(x)dx \right|}{\int_{a}^{b}f(x)dx} $
Remark 1: $\lambda \in \{\pm 1,\pm i\}$.
Remark 2: Integrals are homogeneous, i.e.,
$\forall a\in \mathbb{C} :\int_{a}^{b}a f(x)dx = a \int_{a}^{b}f(x)dx$
Remark 3: $\forall a\in \mathbb{C}: |Re(a)|\leq |a|$
So, we have
$\mathbb{R^+} \ni \left|\int_{a}^{b}f(x)dx\right| = \lambda \int_{a}^{b}f(x)dx = \int_{a}^{b} \lambda f(x)dx = \int_{a}^{b}Re\{\lambda f(x)\}dx \leq \int_{a}^{b}|Re\{\lambda f(x)\}|dx \leq \int_{a}^{b}|\lambda f(x)|dx \leq |\lambda|\int_{a}^{b} |f(x)|dx$
Finally,
$ \left|\int_{a}^{b}f(x)dx\right| \leq \int_{a}^{b}\left|f(x)\right| dx $
- Proving the second inequality:
$\left| \int_{\gamma}f(z)dz \right| = \left| \int_{a}^{b}f(\gamma(t)) \gamma ' (t) dt \right| \leq \int_{a}^{b} \left| f(\gamma(t)) \gamma ' (t)\right| dt = \int_{a}^{b} \left| f(\gamma(t))\right| \left| \gamma ' (t)\right| dt = \int_{\gamma}\left|f(z)\right||dz|$