8-1

i) \int_E f is finite, if and only if each component is finite, i.e. \int_E f_1and \int_E f_2 is finite. This is if and only if f_1,f_2 is finite a.e. in E, hence if and only if \int_E |f|=\int_E \sqrt{f_1^2+f_2^2} \text{ is finite}.
Before that, we will need a trigonometric formula a\cos \alpha+b \sin \alpha=\sqrt(a^2+b^2)\sin(\theta+\alpha)
 , with \theta = \arctan \frac{a}{b}
.
\text{To show the inequality, using the hints we obtain }
|\int_E f|=[(\int_Ef_1)^2+(\int_Ef_2)^2]^\frac{1}{2}=\int_Ef_1\operatorname{cos}\alpha+\int_Ef_2\sin \alpha
, with \alpha=\arctan(\frac{b}{a})
.
Where b=\int_E f_1, a=\int_E f_2
.
Now \begin{align}
\left|\int_E f\right|&=(a\cos \alpha+b \sin \alpha) \\&= Re(a\cos \alpha+b \sin \alpha)\\&=\int_E Re(f_1\cos\alpha+f_2 \sin \alpha)\\ &\leq \int_E |f_1\cos\alpha+f_2 \sin \alpha|\\&\leq \int_E \sqrt(f_1^2+f_2^2)\\&=\int_E|f|.
\end{align}