Suppose $A$ and $B$ are invertible $n \times n$ matrices. Show that there are only finitely many complex numbers $\lambda$ for which $\text{det}( \lambda A + (1-\lambda)B) = 0$. Show that there exists a continuous path $A(t)$ of the form $A(t) = \lambda(t) A + (1 - \lambda (t)) B$ connecting A to B and such that $A(t)$ lies in $GL(n; \mathbb{C})$. Here, $λ(t)$ is a continuous path in the plane with $λ(0) = 0$ and $λ(1) = 1$.
I solved the first part: Since $A$ and $B$ are invertible: $\text{det}( \lambda A + (1-\lambda)B) = 0$ is a non zero polynomial of degree at most n with complex coefficients, so, by the fundamental theorem of algebra, this polynomial have at most n complex zeroes. For the second part I was thinking: first of all $A(t): t \in [0,1] \rightarrow \mathbb{M}_n(\mathbb{C})$, is continuous because $\lambda(t)$ is continuous. Since $\text{det}(A(0)) = \text{det}(B) \neq 0$ and $\text{det}(A(1)) = \text{det}(A) \neq 0$ we can always find in the complex plane a continuous path $\lambda(t)$ with $λ(0) = 0$ and $λ(1) = 1$ that avoids the zero $\forall t \in (0,1)$. Is my proof correct?