Post Closed as "too localized" by Dan Petersen, Deane Yang, Benjamin Steinberg, Willie Wong, Pietro Majer

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Let $A(z)$ be a compact operator on a Hilbert space, depending on a complex parameter $z$. I want to count the number of points where $1+A(z)$ is not invertible and therefore I want to count zeros of $\det (1+A(z))$. Unfortunately $A(z)$ is not of trace class so the determinant does not make sense. However, for some positive integer power $n$ the operator $A(z)^n$ is of trace class. Can I argue that the points $z$ where $1+A(z)^n$ is not invertible include those where $1+A(z)$ is not invertible?

EDIT: I mean $1+(-1)^{n-1}A(z)^n$ rather than $1+A(z)^n$.

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1+A not invertible implies 1+A^n not invertible?

Let $A(z)$ be a compact operator on a Hilbert space, depending on a complex parameter $z$. I want to count the number of points where $1+A(z)$ is not invertible and therefore I want to count zeros of $\det (1+A(z))$. Unfortunately $A(z)$ is not of trace class so the determinant does not make sense. However, for some positive integer power $n$ the operator $A(z)^n$ is of trace class. Can I argue that the points $z$ where $1+A(z)^n$ is not invertible include those where $1+A(z)$ is not invertible?