Decomposition of the absolute value of a complex line integral

Question

I just would like someone to help me understand how $$\Bigg\vert{\int^{b}_{a} g(t) dt} \Bigg\vert = e^{-i \theta} \int_{a}^{b} h(t) dt$$

Answer 1

It is true for any complex number that $z=|z|e^{iArg{z}}$, or equivalently $|z|=e^{-iArg{z}}z$. So here you are just using it for $z=\int_a^b g(t)dt$. Hence $|\int_a^b g(t)dt|=e^{-i\theta}\int_a^b g(t)dt$. Because an integral is linear the last expression is also equal to $\int_a^b e^{-i\theta}g(t)dt$.