determinant identity for invertible finite rank operators

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I am currently reading a paper where the following identity, valid for an invertible finite - rank operator $T \colon \mathscr{H} \to \mathscr{H}$ on a separable Hilbert space, is given: $$ \log \det T = - \frac{d}{dz} \{\text{Tr} \,T^{-z}\}|_{z = 0} $$ Here, $\text{Tr}$ stands for the trace defined on trace class operators, and $z$ is a complex variable. I would like to understand this identity better, in particular I would like to know whether the definition of the complex power on the left hand side is based on a functional integral in general, or whether there are alternative definitions for this. Is there a good source where I could read up on this?

Many thanks!

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In general you would use the holomorphic functional calculus $$ f(T)=\frac1{2\pi i}\,\int_\Gamma \,f(z)\,(z-T)^{-1}\,dz, $$ where $\Gamma$ is a curve in the complex plane with the spectrum of $T$ inside the region it delimits, and $f$ is an analytic function on that region.

Now, if $H$ is finite-dimensional, things are way easier. Then you have $T=SJS^{-1}$ with $J$ the Jordan form, so that it has the eigenvalues $\lambda_1,\ldots,\lambda_n$ in the diagonal. Then $J^{-z}$ has $\lambda_1^{-z},\ldots,\lambda_n^{-z}$ in the diagonal, so $$ \mbox{Tr}(T^{-z})=\mbox{Tr}(J^{-z})=\sum_{j=1}^n\lambda_j^{-z}, $$ and $$ \left.\frac{d}{dz}\right|_{z=0}\,\mbox{Tr}(T^{-z})=\left.\frac{d}{dz}\right|_{z=0}\sum_{j=1}^n\lambda_j^{-z}=-\left.\sum_{j=1}^n\lambda_j^{-z}\log\lambda_j\right|_{z=0}\, =-\sum_{j=1}^n\log\lambda_j\\ =-\log\left(\prod_{j=1}^n\lambda_j\right) =-\log\det T $$