$|\int\limits_a^b f(t)dt| \leq \int\limits_a^b |f(t)|dt$, where $f(t) \in \mathbb C$

Question

Let $f: [a,b] \to \mathbb C, t \to f(t) = \text {Re } f(t) + i\text{ Im } f(t)$. Suppose $f$ is continuous.

Let $\int\limits_a^b f(t)dt = \rho e^{i\theta}$, where $\rho \geq 0$ is the module. Then: $$|\int\limits_a^b f(t)dt| = \rho = e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt = \int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$$

I don't understand 2 steps:

1. $e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt$

2. $\int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$

Answer 1

The first step isn't right, and it should have $+i\theta$ rather than $-i\theta$. This is because

$$\rho = e^{i\theta} \big(\rho e^{-i\theta}\big) = e^{i\theta} \int_a^b f(t) \, dt$$

Now take real parts - since $\rho$ is real, the right hand side is real too.

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The second step is just an application of the fact that the real part is bounded by the modulus, together with the fact that $|e^{i\theta}| = 1$.

Answer 2

For the first step, up to the sign, note that $\rho$ is real. Since it is equal to $e^{-i\theta}\int\limits_a^b f(t)dt$, the last expression is a real number and so $$\rho=e^{-i\theta}\int\limits_a^b f(t)dt= \text{Re }e^{-i\theta}\int\limits_a^b f(t)dt.$$

For the second step, write $z=x-iy$ and note that $$\text{Re }z=x\le |x|\le \sqrt{x^2}\le\sqrt{x^2+y^2}=|z|.$$