Trace of the log of a matrix When Zee computes the Grassmann path integral in Section II.5 of his QFT book, he uses an algebraic step I can't follow.  I follow this part:
\begin{align}
\text{Tr}\ln\!\big[\gamma^5\big(-i\gamma^\mu\partial_\mu-m\big)\gamma^5\big]&=\text{Tr}\!\left\{\ln\gamma^5+\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)+\ln\gamma^5 \right\}\\[4pt]
 &=\text{Tr}\!\left\{\big(\ln\gamma^5+\ln\gamma^5\big)+\ln\!\big(-i\gamma^\mu\partial_\mu-m\big) \right\}\\[4pt]
 &=\text{Tr}\!\left\{\ln\!\big(\gamma^5\big)^{\!2}+\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)\right\}\\[4pt]
 &=\text{Tr}\!\left\{\ln1+\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)\right\}\\[4pt]
 &=\text{Tr} \ln\!\big(-i\gamma^\mu\partial_\mu-m\big) ~~.
\end{align}
Then Zee writes
\begin{align}
     \text{Tr} \ln\!\big(-i\gamma^\mu\partial_\mu-m\big)&=\frac{1}{2}\text{Tr}\!\left\{\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)+\ln\!\big(i\gamma^\mu\partial_\mu-m\big)\right\}\\[4pt]
&=\frac{1}{2}\text{Tr}\ln\!\big(\partial^2+m^2\big)~~,
\end{align}
but I do not see how he got the complex conjugate on the first line.  It appears to me that he did something like
\begin{align}
     \text{Tr} \ln\!\big(-i\gamma^\mu\partial_\mu-m\big) &=\text{Tr}\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)^{2\times\frac{1}{2}} \\[4pt]
      &=\frac{1}{2}\text{Tr} \ln\!\big(-i\gamma^\mu\partial_\mu-m\big)^{2} \\[4pt]
      &=\frac{1}{2}\text{Tr}\!\left\{\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)+\ln\!\big(-i\gamma^\mu\partial_\mu-m\big)\right\}~~,
\end{align}
but I do not see where the complex conjugate of $(-i\gamma^\mu\partial_\mu-m)$ comes from.
 A: The crux of the argument is proving that $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ = $\text{Tr}\ln(-i\gamma^\mu\partial_\mu-m)$. While it might be possible to prove this using the fact that the trace of the log of an operator is the logarithm of the product of its eigenvalues, I wasn't able to do this myself.
Start with $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$, then using the cyclic property of the trace, its linearity, and the expansion of the logarithm of an operator, you get
$$
\text{Tr}\ln(\mathbb{I}_4\ (i\gamma^\mu\partial_\mu-m)) \\
= \text{Tr}\ln((\gamma^5)^2 \ (i\gamma^\mu\partial_\mu-m)) \\
= \text{Tr}\ln(\gamma^5(i\gamma^\mu\partial_\mu-m)\gamma^5)
$$
Then using the fact that $\{\gamma^5, \gamma^\mu\} = 0$, the $\gamma^5$'s act on the $\gamma^\mu$ by conjugation to yield $-\gamma^\mu$, while $m$ commutes with the $\gamma^5$, yielding
$$
\text{Tr}\ln(-i\gamma^\mu\partial_\mu-m)
$$
So, using the linearity of the trace again, we can break up $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ into $\frac12 \text{Tr} [ \ln(i\gamma^\mu\partial_\mu-m) + \ln(i\gamma^\mu\partial_\mu-m)]$ and rewrite the second one using the previous identity.
