Finding pairs of complex numbers with a certain property

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Finding pairs of complex numbers with a certain property

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Source: Romania NMO 2025 10.4

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#1 h Apr 2, 2025, 1:22 PM

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Find all pairs of complex numbers $(z,w) \in \mathbb{C}^2$ such that the relation $|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1$ holds for all positive integers $n$ .

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#2 h Apr 3, 2025, 5:20 AM

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It's easy to see $z,w \neq 0$ . Wlog assume that $|z| \ge |w|>0$ and let $t=w/z$ , then $|z|^{2n} \cdot |1+t^n+t^{2n}|=2^{2n}+2^n+1$ . We have $|1+t^n+t^{2n}| \le 3$ . If $|t|<1$ then $|1+t^n+t^{2n}| \ge 1-|t|^n-|t|^{2n}>1/2$ for all large $n$ ; if $|t|=1$ , we can also choose large $n$ such that $\arg(t^n) \in (-\pi/4,\pi/4)$ , so $|1+t^n+t^{2n}| \ge 1$ . Hence for both cases, by choosing the appropriate large $n$ we get that $|z|=2$ .

Now the equation becomes $|1+t^n+t^{2n}|=1+2^{-n}+2^{-2n}$ . Clearly we have $|1+t^n+t^{2n}| \le 1+|t|^n+|t|^{2n}$ which implies $|t| \ge 1/2$ . If $|t|=1$ , for any $\theta>0$ we can choose large $n$ such that $\arg(t^n) \in (-\theta,\theta)$ , making $|1+t^n+t^{2n}|$ sufficiently close to 3, which is a contradiction. Now if $|t| \in (1/2,1)$ , then by choosing large $n$ such that $\arg(t^n) \in (-\theta,\theta)$ , we can make $|1+t^n+t^{2n}| \ge |1+t^n|-|t|^{2n} \ge 1+\tfrac{1}{2}\cos\theta \cdot |t|^{-n}>1+2^{-n}+2^{-2n}$ , a contradiction. Hence $|t|=1/2$ , and thus $1,t^n,t^{2n}$ must be of the same direction for the triangle inequality to attain equality, deducing $t=1/2$ . Therefore the solution is all $(z,w)$ such that $|w|=1$ and $z=2w$ or $|z|=1$ and $w=2z$ .

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