Let $n\geq2$ and let $\{z_1,\dots,z_n\}$ be a set of complex numbers.
Is there a condition on the $z_i$'s such that $$\sum_{i=1}^n z_i=\prod_{i=1}^n z_i$$ is identically true?
For $n=2$ the condition that $$a+b=ab$$ just becomes $b=\frac{a}{a-1}$, for $a-1\neq0$, but is there an inductive approach to solving the general problem?
My reason of investigating this is to learn about the class of linear operators $A$ with the property that $\det(A)=\mathrm{tr}(A)$, in which case the above condition is the condition on the eigenvalues.
Thanks in advance :)
Define
$$r=\sum_{j=1}^{n-1} z_j, \quad s=\prod_{j=1}^{n-1} z_j$$
(Note the $n-1$ as the upper bounds rather than $n$.) Then your equation is equivalent to
$$r+z_n=sz_n$$
which is solved by
$$z_n=\begin{cases} \frac r{s-1} & \text{if $s\ne 1$} \\[2ex] \text{anything} & \text{if $s=1, r=0$} \end{cases}$$
(There is no solution if $s=1,r\ne 0$.) Therefore your desired condition is just on any one of the $z_k$'s, except for one special case where your equation cannot be satisfied.