$\newcommand\la\lambda$If (all the entries $a_{ij}(z)$ of) the matrix $A(z)$ are continuous in $z$, then, yes, $\la_z$ is continuous in $z$. 

Indeed, take any complex $z$ and any sequence $(z_k)$ converging to $z$. Then, according to (say) [Sections 5.1 and 5.2 of Chapter II][1], the set of eigenvalues of $A(z_k)$ converges to the set of eigenvalues of $A(z)$ in the following sense: for some enumeration $(\la_1(z),\dots,\la_n(z))$ of the eigenvalues of $A(z)$ and some enumerations $(\la_1(z_k),\dots,\la_1(z_k))$ of the eigenvalues of $A(z_k)$ one has 
$$\max(|\la_1(z_k)-\la_1(z)|,\dots,|\la_n(z_k)-\la_n(z)|)\to0$$ 
(as $k\to\infty$). It follows that 
$$\la_{z_k}=\max(|\la_1(z_k)|,\dots,|\la_n(z_k)|)
\to(|\la_1(z)|,\dots,|\la_n(z)|)=\la_z.\quad\Box$$

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Clearly, the condition that $A(z)$ be continuous in $z$ cannot be dropped. For example, consider the case when $a_{11}(z)$ is not continuous in $z$ whereas $$a_{ij}(z)=0$ for all $z$ if $(i,j)\ne(1,1)$. 


  [1]: https://link.springer.com/book/10.1007/978-3-642-66282-9