I was reading this expository paper by Chris Blair on finding the irreducible representations of $S_5$. In Section 2.5 of the paper, the author makes an elementary argument using eigenvalues to determine the possible characters of the 5-cycle, 4-cycle, and 3-cycle in an arbitrary irreducible 5-dimensional representation of $S_5$. I think that I understand the arguments made for the 5-cycle and 3-cycle, but I am quite confused about the 4-cycle. I attach the text below for easier reference [I( , ) denotes the Hermitian Inner Product on the Space of Class Functions]:
Similarly consider a 4 -cycle $(a b c d)$ with eigenvalues $\pm 1, \pm i$. As $(a b c d)^{3}$ is also a 4 -cycle, if $i$ occurs as an eigenvalue then so does $i^{3}=-i$. Hence $\pm i$ occur together (or not at all) and will cancel out when we take the trace, leaving an odd number of $\pm 1$ pairs. The two possibilities for $\chi_{\varphi}((a b c d))$ are $\pm 3, \pm 1$ - we again see that the former contributes too much to $I(\varphi, \varphi)$, hence $\chi_{\varphi}((a b c d))=\pm 1$. We choose the plus option - the negative possibility then corresponds to $\varepsilon \varphi$.
I think I understand why $\pm i$ must occur together (or not at all), and that they will cancel out in the trace, but I don't understand why this fact implies that what is left is "an odd number of $\pm 1$ pairs". If $\pm i$ do not occur at all, is it not possible that the trace is $\pm 5$? (From my understanding, we will ultimately rule these out as contributing too much to $I(\varphi, \varphi)$, but at this point in the argument I don't see why this is not a possiblity).
Thank you for your help! (This is my first post so please let me know if I'm doing anything I shouldn't be doing.)