I have tried to apply school methods here and I have troubles with case $0\in D$. Does anybody manage this case? BTW what results on this problem during TST?

you must probably mean you have problems in the case $0 \notin D$, right? the only student that solved the problem (but not completely) was iura . also andreis had some ideas about how to proceede, but did not complete his paper with the details.

I mean exactly what I said: problems with case $0\in D$. Maybe I don't see some obvious things.

Hi myth, your avatar looks smart and relaxed. I really could imagine you studying problems and looking like him. But kind of younger...

Hi Orlando! And I like your mysterious avatar!

And what about solution? Does anybody know it?

I repeat my question: What about solution? Does anybody know it?

Where did you find this lemma?

I had this idea about taking the disc of minimal radius which covers the $n$ points and proving that we can modify the positions of the points s.t. the product of the complex numbers representing them remains the same but the radius of the disc decreases. Then we consider the infimum of the set of such radii and show that there must be some disc with strictly inferior radius as long as we assume the infimum to be $>0$, but I don't know if it works (and I got $0$ pts during the examination , although I didn't use this ).

that's what I did: If any of the $z_i=0$ then we can take $z=0$. If all the $z_i\neq 0$ we can take off the disk a segment math=inline]$0,w$[/math $=\{tw\,/\,0\leq t\leq 1\}$ with $|w|=1$ and $z_1,\ldots,z_n\in D-[0,w]$. Now we can define an holomorfic logarithm $ln$ in $D-[0,w]$ so \[z=exp\left(\frac{ln z_1+\ldots+ln z_n}{n}\right)\] works.

Suppose the $z_j$ are all nonzero. First we address the $n=2$ case; WLOG assume $z_1z_2=1$, so we want to show that for any $z\ne0$ and disk $D$ containing $z,1/z$, one of $\pm1$ must lie in $D$. Note that any such $D$ must contain a smaller $D'$ with $z,1/z$ on its boundary. If $z\in\mathbb{R}$, $\operatorname{sgn}(z)\in\{1,-1\}$ must always lie in $D'\subseteq D$. Otherwise, the identity $\frac{z-1}{1/z-1}\bigg{/}\frac{z+1}{1/z+1} = -1$ tells us that $z,1,1/z,-1$ form a harmonic quadrilateral, whence the segments from $z$ to $1/z$ and $-1$ to $1$ must intersect. In particular, if $D'$ is not the circumcircle of $z,1,1/z,-1$, it must contain exactly one of $1,-1$ in its interior, so we're done. Now suppose $n\ge 3$ and WLOG assume $z_1z_2\cdots z_n = 1$. The key is that for any two indices $j,k$, the $n=2$ case gives us a geometric mean $z_{j,k}$ of $z_j,z_k$ with $z_{j,k}\in D$ and $|z_{j,k}| = \sqrt{|z_j|\cdot |z_k|}$. Call this $z_j,z_k\mapsto z_{j,k},z_{j,k}$ operation $T_{j,k}$; then there exists an infinite sequence* $(j_1,k_1),(j_2,k_2),\ldots$ such that for any $\epsilon>0$, there exists a positive integer $N(\epsilon)$ such that for all positive integers $N\ge N(\epsilon)$ and indices $j$, $2^{-\epsilon} \le |z_j^{(N)}| \le 2^{\epsilon}$. (Here ${z_j^{(N)} = (T_{j_N,k_N}\circ T_{j_{N-1},k_{N-1}}\circ \cdots \circ T_{j_1,k_1}}) (z_j)$.) To finish, let $c\in\mathbb{C}$ and $r\ge0$ denote the center and radius of $D$, respectively. For each $\epsilon>0$ and $N\ge N(\epsilon)$, we have $z_1^{(N)}\cdots z_n^{(N)} = 1$, so there must exist two roots of unity $\omega_1 = \omega_1^{(N)},\omega_2 = \omega_2^{(N)}$ and indices $j,k$ such that (the arguments of) $z_j^{(N)},\omega_1,z_k^{(N)},\omega_2$ run counterclockwise in that order. But if we let $z'_j = z_j^{(N)}/|z_j^{(N)}|$ and $z'_k = z_k^{(N)}/|z_k^{(N)}|$, then \[ |c-z'_j| \le |c-z_j^{(N)}| + |z_j^{(N)} - z'_j| \le r + \lvert |z_j^{(N)}| - 1\rvert \le r + (2^\epsilon-1). \]Yet $z'_j,\omega_1,z'_k,\omega_2$ lie in that order on the unit circle, so a similar argument to that of the $n=2$ case shows that at least one of $|c-\omega_1|,|c-\omega_2|$ is bounded above by $r+(2^\epsilon - 1)$ ("close" to $D$). Hence for any $\epsilon>0$, there exists an $n$th root of unity at most distance $2^\epsilon - 1$ from $D$. Since there are only finitely many such roots and $\lim_{\epsilon\to0} 2^\epsilon - 1 = 0$, one of the $n$th roots of unity must lie in $D$, as desired. *To prove this, let $a_j = \log_2 |z_j|$ and note that $\sum a_j = 0$. If $|a_j|\in [\epsilon,2\epsilon)$, there must then exist $a_k$ of opposite sign from $a_j$, so $|(a_j+a_k)/2| < \epsilon$ means $T_{j,k}$ reduces the number of $l$ with $|a_l|\in [\epsilon,2\epsilon)$ by at least one. Repeating this proves the desired claim.

The following post by Andrew was lost in the crash a while ago, but I found it in an online PDF (attached). (Well, I've edited it a bit to make it slightly more natural---but perhaps only from the perspective of coming up with the problem from scratch, rather than purely solving the problem.) Suppose monic $f(x) = \prod (x-t_k)$ of degree $n$ has complex roots $t_1,\ldots,t_n$, and all $t_i$ lie (strictly) outside some closed disk $D$ (call the boundary, a circle, $C$) in the complex plane. Let $u$ be an arbitrary point in $D$. Then along the lines of the Gauss-Lucas theorem convexity argument, note that an inversion $I$ (say of radius $1$) about $u$ sends $t_1,\ldots,t_n$ to points $I(t_1),\ldots,I(t_n)$ (strictly) inside the image $I(C)$ (a circle). In particular, $\frac{I(t_1)+\cdots+I(t_n)}{n}$ lies inside $I(C)$ by convexity, so it equals $I(\alpha)$ for some $\alpha$ strictly outside $D$. But of course, $I(z) = \frac{z-u}{|z-u|^2} = \frac{1}{\overline{z}-\overline{u}}$, so we've shown that if $\alpha$ satisfies $\frac{1}{n} \frac{-f'(u)}{f(u)} = \frac{1}{n} \sum \overline{I(t_k)} = \overline{I(\alpha)} = \frac{1}{\alpha - u}$, then $\alpha\notin D$. More powerfully, consider the contrapositive statement: if $\alpha\in D$, then all solutions $u$ to $\frac{1}{n} \frac{-f'(u)}{f(u)} = \frac{1}{\alpha - u}$ must have $u\notin D$. --- Applied to our problem, go by contradiction with $n$ WLOG minimal (so $n\ge2$), and take $f(x) = x^n - z_1\cdots z_n$. Then for all $\alpha\in D$, the roots of $g(x) := n f(x) - (x-\alpha) f'(x) = n\alpha x^{n-1} - nz_1\cdots z_n$ must all lie strictly outside $D$. But then taking $\alpha = z_n$, say, contradicts the $n-1$ case, so we're done.

Attachments:
selection2004sol.pdf (70kb)

Oh wow, see Polya/Szego, Problems in analysis II, Part Five. The Location of Zeros, Chapter 2. The Geometry of the Complex PllDe IDd the Zeros of Polynomials, Section 3. Derivative of a Polynomial with respect to a Point. (Problem 134 in my edition basically corresponds to the lemma above... this book is ridiculous...)