I am studying Lie groups on my own via Brian Hall's Lie groups, Lie algebras, and representations. I am currently stuck on the first part of exercise 12 in chapter 1. The question is as follows:
Suppose $A$ and $B$ are invertible $n\times n$ matrices. Show that there are only finitely many complex numbers $z$ for which $\mathrm{det}[zA+(1-z)B]=0$.
I thought I should add some context to how I have tried to approach this problem and where I am stuck. I know that the determinant of a matrix is zero iff it has zero for at least one eigenvalue. Therefore, the problem reduces to showing that $zA+(1-z)B$ has zero for an eigenvalue for only finitely many complex numbers $z$. Then I know that for some vector $v$ it follows that $Av=(1-z)/z B$. At this point I am unsure as to how to continue.
A better approach is to observe that ${\rm det}[zA + (1 - z)B]$ is a polynomial of degree at most $n$ in the variable $z$. By the fundamental theorem of algebra, this polynomial can have at most $n$ complex zeroes.