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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.

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