Let, $G$ be a finite abelian group of order $n$. So there are exactyly $n$ characters of the group $G$. For each character $f$ we have $|f(a)|=1$.Hence the reciprocal $\displaystyle \frac{1}{f(a)}$ is equal to the complex conjugate $\overline{f(a)}$. Thus the function $\bar f$ defined by $\bar{f}(a)=\overline{f(a)}$ is also a character of $G$. Moreover we have, $\displaystyle \bar f(a)=\frac{1}{f(a)}=f(a^{-1})$ ofr every $a\in G$.
My question is on the first line. How $|f(a)|=1$ for each character $f$ of $G$ ?
Since $f$ is a character and $G$ has order $n$, we have $$ 1=f(e)=f(a^n)=f(a)^n$$ where $e$ is the identity element of $G$.
Therefore $f(a)$ is an $n$th root of unity, so in particular $|f(a)|=1$.