$\operatorname{Tr}\det H$ of a $2\times 2$ block matrix $H$

Question

Let $H$: $$ H=\left(\begin{array}{cc}{0} & {\phi-z} \\ {\phi^{\dagger}-z^{*}} & {0}\end{array}\right) $$ where $\phi$ is a $N\times N$ matrix.

In their paper (https://arxiv.org/abs/cond-mat/9703087), Feinberg and Zee tell us that: $$ \left\langle\operatorname{tr}_{(N)} \log (z-\phi)\left(z^{*}-\phi^{\dagger}\right)\right\rangle=\left\langle\operatorname{tr}_{(2 N)} \log H\right\rangle- i \pi N^{2}$$

How does one show this?

My attempt: $\operatorname{Tr}\log H=\log \det H=\log \det\left(\mathbf{0}-(z-\phi)\left(z^{*}-\phi^{\dagger}\right)\right)=\log\left[(-1)^N\det\left(z-\phi\right)\left(z^{*}-\phi^{\dagger}\right)\right]$

$\implies \operatorname{Tr}\log H=N\log(\mathrm{i}^2)+\log\det(z-\phi)\left(z^{*}-\phi^{\dagger}\right)=N\mathrm{i}\pi+ \operatorname{Tr}\log(z-\phi)\left(z^{*}-\phi^{\dagger}\right)$

$$\implies \operatorname{Tr}\log(z-\phi)\left(z^{*}-\phi^{\dagger}\right)=\operatorname{Tr}\log H-\mathrm{i}\pi N $$

What am I missing?

Answer 1

OP's calulations are correct. The extra factor of $N$ in Ref. 1 is either

1. a normalization $\left\langle 1\right\rangle=N$ of the average,

2. or an error.

The fact that the extra factor of $N$ does not appear in Ref. 2 by the same author suggests the 2nd possibility.

References:

1. J. Feinberg & A. Zee, arXiv:cond-mat/9703087 ; paragraph above eq. (3.1).

2. J. Feinberg, arXiv:cond-mat/0603622; paragraph above eq. (12).