Gauß sum and primitive character

Question

I am working with Daniel Marcus "Number Field" Book. And I have a question to the following Lemma:
$$\tau_k(\chi)=\left\{\begin{array}{ll} \bar\chi(k)\tau(\chi), & \textrm{if }(k,m)=1 \\ 0, & \textrm{if }(k,m)>1\end{array}\right.$$ With $\chi$ a primitive character modulo $m$. For the proof of the case $(k,m)=1$ he writes:
as $a$ runs through $(\mathbb{Z}/m\mathbb{Z})^*$, so does $ak$ if $(k,m)=1$.
I understand this argument, but not why $\tau_k(\chi)=\bar\chi(k)\tau(\chi)$ follows.

If I consider $\tau_k(\chi) = \sum\limits_{a\in(\mathbb{Z}/m\mathbb{Z})^*}\chi(a)e^{2\pi iak/m}$. Then By the argument above I can say:
$\sum\limits_{a\in(\mathbb{Z}/m\mathbb{Z})^*}\chi(a)e^{2\pi iak/m} = \sum\limits_{a\in(\mathbb{Z}/m\mathbb{Z})^*}\chi(ak)e^{2\pi iak/m} = \chi(k)\sum\limits_{a\in(\mathbb{Z}/m\mathbb{Z})^*}\chi(a)e^{2\pi iak/m} = \chi(k)\tau_k(\chi)$.

How does he get $\bar\chi(k)$ instead of $\chi(k)$. And what exactly is $\bar\chi(k)$?

Answer 1

Note that $\;\chi(k)^{-1}\chi(k)=\overline{\chi}(k)\chi(k)=|\chi(k)|^2=1\;$ , so

$${}$$

$$\tau_k(\chi)=\sum_{a\in\Bbb Z_m^*}\chi(a)e^{2\pi iak/m}=\sum_{a\in\Bbb Z_m^*}\chi(k)^{-1}\chi(ak)e^{2\pi iak/m}=$$

$${}$$

$$=\overline{\chi}(k)\sum_{ak=:b\in\Bbb Z_m^*}\chi(b)e^{2\pi ib/m}=\overline{\chi}(k)\tau$$

Ah, I still remember this stuff!