This is Isaac's exercise 3.16.
Let $G$ be a group with odd order. Show that if $\chi \in Irr(G)$ is not principle then $\chi$ can't equal to its conjugate.
my try: Suppose $\chi = \bar \chi$
By the orthogonality $[\chi,\bar\chi]=1$
$$\sum_{g\in G} \chi^{2}=|G|$$
I know that the degree divides the the group's order, so
$$\chi(1) \mid \sum_{g\in G-1} \chi^{2}$$
I said $\chi(1)$ is odd ($\chi(1) \mid |G| $ odd integer)
Also, the summation is an even integer
I don't feel like I am going any closer
Let $\chi \in Irr(G)$, with $\chi \neq 1_G$ (the trivial character). Apply the orthogonality relation to $\chi$ and $1_G$, then $$\sum_{x \in G}\chi(x)=0.$$ Now assume that $\chi$ is real-valued. Then $\chi(x)=\overline{\chi(x)}$ (complex conjugate) for all $x \in G$. From the orthogonality formula above one has $$\chi(1)=-\sum_{x \in G-\{1\}}\chi(x).$$ But $\chi(1) \mid |G|$, so the degree of $\chi$ is odd. The oddness of $|G|$ also implies that for all $x \in G-\{1\}$, $x\neq x^{-1}$, so $$\sum_{x \in G-\{1\}}\chi(x)=\sum \chi(x)+\chi(x^{-1}),$$ where the last sum is taken over a set of disjoint pairs $\{x,x^{-1}\}$. Now $\chi(x)$ is the sum of the eigenvalues of a matrix representing $x$, while $\chi(x^{-1})$ is the sum of the eigenvalues of the inverse of that same matrix. The eigenvalues of the matrix are roots of unity and for a root of unity $\zeta$, one has $\zeta^{-1}=\overline{\zeta}$. Since $\chi$ is real-valued one must have $\chi(x)=\chi(x^{-1})$. It follows that $$\frac{1}{2}\chi(1)=-\sum_{x}\chi(x).$$ We now have a contradiction: the left hand side is half an odd positive integer, while the right hand side is an algebraic integer.