I'm following the proof of the Fourier expansion of primitive Dirichlet character from theorem 8.20 in Apostol's Introduction to Analytic Number Theory. Here is the complete proof:
I do not see the last equality, i.e., why $\sum_{n = 1}^k \bar{\chi}(-n) e^{2\pi i m n/k} = \sum_{n = 1}^k \bar{\chi}(n) e^{-2\pi i m n/k}$.
Thanks very much for your help.
You want to show that $$ \sum_{n = 1}^k \bar{\chi}(-n) e^{2\pi i m n/k} = \sum_{n = 1}^k \bar{\chi}(n) e^{-2\pi i m n/k}. $$ The short summary is that this is a change of variables in the sum, where $n \mapsto -n$.
For a more detailed explanation, let us break this equality into a few steps. Firstly, note that $$ \sum_{n = 1}^k \overline{\chi}(-n) e^{2\pi i m n / k} = \sum_{n = 1}^k \overline{\chi}(k-n) e^{2\pi i m (k-n) / k}, $$ since both $\chi(x)$ and $e^{2\pi i x / k}$ are periodic functions with period $k$. Let $N = k - n$. Then $$ \sum_{n = 1}^k \overline{\chi}(k-n) e^{2\pi i m (k-n) / k} = \sum_{N = k-1}^0 \overline{\chi}(N) e^{2\pi i m N / k} = \sum_{N = 0}^{k-1} \overline{\chi}(N) e^{2\pi i m N / k}. $$ As $\chi(0) = \chi(k) = 0$, this last sum is equal to $$ \sum_{N = 1}^{k} \overline{\chi}(N) e^{2\pi i m N / k}. $$ This is precisely the desired sum.