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Algebra Difficulty 4.7 AIME Prove it United States

Problem:

Solve for all complex numbers zz such that z4+4z2+6=zz^{4} + 4z^{2} + 6 = z.

Solution

Solution:

Rewrite the given equation as (z2+2)2+2=z\left(z^{2} + 2\right)^{2} + 2 = z. Observe that a solution to z2+2=zz^{2} + 2 = z is a solution of the quartic by substitution of the left hand side into itself. This gives z=1±i72z = \frac{1 \pm i \sqrt{7}}{2}. But now, we know that z2z+2z^{2} - z + 2 divides into (z2+2)2z+2=z4+4z2z+6\left(z^{2} + 2\right)^{2} - z + 2 = z^{4} + 4z^{2} - z + 6. Factoring it out, we obtain (z2z+2)(z2+z+3)=z4+4z2z+6\left(z^{2} - z + 2\right)\left(z^{2} + z + 3\right) = z^{4} + 4z^{2} - z + 6. Finally, the second term leads to the solutions z=1±i112z = \frac{-1 \pm i \sqrt{11}}{2}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.