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## Geometric mean for n complex numbers

Let $D$ be a closed disc in the complex plane. Prove that for all positive integers $n$, and for all complex numbers $z_1,z_2,\ldots,z_n\in D$  there exists a $z\in D$ such that $z^n = z_1\cdot z_2\cdots z_n$.

Romanian TST 2004

Proof: Using this lemma the problem becomes very easy, if we choose the right polynomial. First, if one of the $z_i=0$ the problem is solved choosing $z=0$. Suppose now that $z_1...z_n \neq 0$. We argue by contradiction. Suppose $z^n\neq z_1...z_n,\ \forall z \in D$. Take the polynomial $f(z)=z^n-z_1z_2...z_n$, which doesn’t have any of its roots in $D$, by our assumption. Then using the lemma for $x=z_1$ we see that the roots of the polynomial $g(z)=(z_1-z)\cdot n \cdot z^{n-1}-n\cdot (z^n-z_1...z_n)=nz_1(z^{n-1}-z_2...z_n)$ are not in $D$. Therefore, the roots of $f_1(z)=z^{n-1}-z_2...z_n$ are not in $D$. This means we have reduced the problem for $n$ to the problem for $n-1$. Doing this until we reach $n=1$ we see that the polynomial $h(z)=z-z_n$ does not have its roots in $D$. But the only root of this polynomial is $z_n$, which is in $D$. Contradiction.

Therefore, our assumption was false and there exists a root $z$ of $f(z)=0$ in $D$.

The solution presented here is a lot shorter than the official solution in the contest. Moreover, there was only one participant who completely solved the problem.