existence of a complex sequence

Question

Let $z_n$ be a sequence of nonzero complex numbers such that $z_{n+1}=z_n^2+z_n$ for all $n\in\mathbb{Z}$.

Can such a sequence satisfy $\lim_{n\to +\infty}z_n=\lim_{n\to -\infty}z_n=0$ ?

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A friend from the university showed me this question saying it's a student competition problem from the 90's. None of us and our colleagues was able to solve it, nobody had even an idea how to start the solution.

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It is NOT a duplicate. There is a big difference between real and complex numbers and sequences, especially when one wants to use the fact that being bounded and monotonic implies convergence, which was the main observation in the linked question!

Answer 1

NOT AN ANSWER- JUST SOME COMPUTATIONS THAT WERE TOO LONG FOR COMMENT.

Let's compute the inverse recurrence relation using the general solution to a quadratic equation.

$$z_{n+1}=z_n^2+z_n$$

So:

$$z_{n-1}= -\frac{1}{2} \pm \frac{\sqrt{1 + 4z_n}}{2}$$

Since we know the sequence converges to the RHS for any real $-1 < z_n < 1$, we could try to solve this problem by just showing the LHS convergence starting at some $-1 < z_n < 1$ and following the above recurrence relation which decrements $n$.

Now, if we have a converging sequence to the LHS, then $z_n$ will tend towards $0$ and so $\frac{\sqrt{1 + 4z_n}}{2}$ will tend towards $\frac12$ and so beyond a certain point, as we go left, we will no longer be able to choose the minus side of the plus-or-minus sign in the above recurrence relation. In other words, for any arbitrarily small $\epsilon$, we'll need to reach a point $m$ such that $|z_m| < \epsilon$ and forall $i < m$:

$$z_{i}= -\frac{1}{2} + \frac{\sqrt{1 + 4z_{i+1}}}{2}$$

We'd also be forced to have $|z_i| < \epsilon$ for all such $i$ all the way to $-\infty$.

Answer 2

Let $z_0:=-1/2\;$ and $\;z_{n+1}:=z_n^2+z_n.\;$ It is easy to see that $z_n$ decreases monotonically to $0$. Let $z_{n-1}:=(-1+\sqrt{1+4z_n})/2,\;$ taking the principal branch of square root. Observe that with these definitions, $z_{n+1}=z_n^2+z_n$ for all $n\in\mathbb{Z}.$ The first few values for $n<0$ are $z_{-1}=-1/2+i/2,\;$ $z_{-2}=-0.107+0.636i,\;$ $z_{-3}=0.130+0.504i.$ The result to prove is that if $n<0,\;$ then $|z_{n-1}|<|z_n|.\;$ The proof is not much different than the proof for $n>0.\;$ It depends on finding the approximation $|z_n|\approx 1/|n|\;$ for $|n|$ big enough. In fact, $z_{-n}\approx-z_n\;$ for large $|n|$.