Finitely generated $\mathbb{C}$-algebra, uncountably many $\lambda \in \mathbb{C}$. Let $A$ be a finitely generated $\mathbb{C}$-algebra and let $a \in A$ be a nonalgebraic element. My question is, are there uncountably many $\lambda \in \mathbb{C}$ such that the element $a - \lambda$ is not a zero divisor but, at the same time, it is not invertible?
 A: There are two possible ways to interpret your question, from what I can tell. First, I will interpret algebraic to mean algebraic over $\mathbf{Q}$. Then, the answer to your question is: maybe. 
Here are two examples of $A$ and $a$ for which the answer is no: If $A=\mathbf{C}$ and $a=\pi$ then evidently $a-\lambda$ is invertible for all $\lambda \neq \pi$, so the condition doesn't hold. Likewise, if $A=\mathbf{C}[x]/(x^2)$ and $a=x$ then $a-\lambda$ is invertible for all $\lambda \neq 0$, so again the condition doesn't hold. Evidently something similar happens any time your algebra contains a non-trivial nilpotent. 
Here is an example for which the answer is yes: If $A=\mathbf{C}[x]$ is the ring of polynomials in one variable and $a=x$ then $a-\lambda$ is neither invertible nor a zero-divisor for any $\lambda \in \mathbf{C}$. 
There is a second interpretation for your question (related to the Nullstellensatz) as follows: suppose by algebraic you mean satisfying a polynomial equation with coefficients in $\mathbf{C}$ (obviously this condition will be most interesting when $A$ is not necessarily commutative). Then the answer to your question is yes, and in somewhat more generality. Observe that a finitely generated algebra is of at most countable dimension.
Suppose $A$ is an associative (but not necessarily commutative) $\mathbf{C}$-algebra of countable dimension over $\mathbf{C}$ and $a \in A$. If the set of complex numbers $\lambda$ with $a-\lambda$ invertible is uncountable, then the set of elements $(a-\lambda)^{-1} \in A$ must be linearly dependent, whence $a$ is algebraic over $\mathbf{C}$. It follows that for non-algebraic $a \in A$ the set of complex numbers $\lambda$ so that $a-\lambda$ is not invertible is uncountable. 
Now for zero-divisors: for any $a \in A$ and $\lambda \in \mathbf{C}$, if there is non-zero $b \in A$ with $(a-\lambda)b=0$ then the $\lambda$-eigenspace for left-multiplication by $a$ is non-zero. It follows that there are at most countably many $\lambda$'s such that $a-\lambda$ is a left-zero divisor; by symmetry there are at most countably many $\lambda$'s such that $a-\lambda$ is a zero divisor.
