I was trying to follow a derivation on the complex fourier series, but I am a bit confused at one particular step. In the following video https://www.youtube.com/watch?v=Ft5iyapkSqM, at 6:30, the proffessor tries to merge the sums together. I am confused as to how he does this.
From 6:30, we have that $f(\alpha) = a_0 + \sum _{1} ^{\infty}\frac{a_n-ib_n}{2}e^{in\alpha}+\sum _{-\infty} ^{-1}\frac{a_{-n}+ib_{-n}}{2}e^{in\alpha}$. He then labels the expression $\frac{a_n-ib_n}{2}$ as $c_n$ and $a_0 $ as $c_0$. This labelling reduces the above summation to the following: $f(\alpha) = \sum _{0} ^{\infty}c_n e^{in\alpha}+\sum _{-\infty} ^{-1}\frac{a_{-n}+ib_{-n}}{2}e^{in\alpha}$. I do not see how he manages to equate the expression $\frac{a_{-n}+ib_{-n}}{2}$ as $c_{n}$ for negative n, since this expression is the conjugate of $c_n$ for negative n. Can someone explain how the summation was reduced from the first expression to the known: $f(\alpha) = \sum _{-\infty} ^{\infty}c_ne^{in\alpha}$.
It's not clear to me what the question is. It may be helpful to note that no, $a-ib$ is not the conjugate of $a+ib$ unless $a$ and $b$ are real. Now, if $f$ is real-valued then $a_n$ and $b_n$ are real. So the above appears to show that $c_{-n}=\overline{c_n}$. This is no problem, since the $c_n$ do satisfy that (if $f$ is real-valued).