Why is this integral equality true? $\int\limits_{-\pi}^{\pi}|a+be^{it}|dt=\int\limits_{-\pi}^{\pi}\left||a|+|b|e^{it}\right|dt$

Question

I was reading an article, and while proving a proposition, they state that if $a,b\in\mathbb{C}$, then due to periodicity we have $$\int\limits_{-\pi}^{\pi}|a+be^{it}|dt=\int\limits_{-\pi}^{\pi}\left||a|+|b|e^{it}\right|dt$$ I don't see it clearly, and I'm trying to see why is this true.

The only thing that comes to me is writting $$|a+be^{it}|=\left||a|e^{it_1}+|b|e^{it_2}e^{it}\right|$$ but I don't know what to do from here.

Answer 1

Hints: Following on from what you have done, firstly, try and convince yourself that

$$\int_{-\pi}^{\pi}\left||a|e^{it_1} + |b|e^{i(t\color{red}{+t_2})}\right|\, dt = \int_{-\pi}^{\pi}\left| |a|e^{it_1} + |b| e^{it}\right|\, dt.$$

(This uses periodicity.)

Secondly, factor out $e^{it_1}$ from the modulus of the right-hand side above and use periodicity similarly to above to note that

$$\int_{-\pi}^{\pi}\left||a|+ |b|e^{i(t\color{red}{-t_1})}\right|\, dt = \int_{-\pi}^{\pi}\left| |a|+ |b| e^{it}\right|\, dt.$$