1. ## complex foorier question

$a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\\$
$\frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi}\overline{f(t)}\overline{e^{int}}dt\\$
$=\frac{e^{in\pi}}{2\pi}\int_{\pi}^{-\pi}e^{-i(-n)t}=e^{in\pi}\overline{a_{-n}}$
why there is $\overline{a_{-n}}$??
by the first formula it supposed to be only $a_{-n}$

i understand how they removed the conjugat of exponenet
but i cant see how they removed it from the f part

2. The formula you gave for $a_n$ was
$a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! f(t)e^{-int} \, dt$, but they tell you what the value of
$\frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)}\overline{e^{int}} \,dt$. Now you have to convert the original formula of $a_n$ to look like the integral you know the value of, i.e.,
$a_{-n} = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! f(t)e^{-i(-n)t} \, dt$
$\overline{a_{-n}} = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)e^{int}} \, dt$
$e^{in\pi}\overline{a_{-n}} = \frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)e^{int}} \, dt$.