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Algebra Difficulty 5.7 AIME, harder Find the answer

The sequence (zn)\left(z_{n}\right) of complex numbers satisfies the following properties: z1z_{1} and z2z_{2} are not real. zn+2=zn+12znz_{n+2}=z_{n+1}^{2} z_{n} for all integers n1n \geq 1. zn+3zn2\frac{z_{n+3}}{z_{n}^{2}} is real for all integers n1n \geq 1. z3z4=z4z5=2\left|\frac{z_{3}}{z_{4}}\right|=\left|\frac{z_{4}}{z_{5}}\right|=2 Find the product of all possible values of z1z_{1}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

All complex numbers can be expressed as r(cosθ+isinθ)=reiθr(\cos \theta+i \sin \theta)=r e^{i \theta}. Let znz_{n} be rneiθnr_{n} e^{i \theta_{n}}. zn+3zn2=zn+22zn+1zn2=zn+15zn2zn2=zn+15\frac{z_{n+3}}{z_{n}^{2}}=\frac{z_{n+2}^{2} z_{n+1}}{z_{n}^{2}}=\frac{z_{n+1}^{5} z_{n}^{2}}{z_{n}^{2}}=z_{n+1}^{5} is real for all n1n \geq 1, so θn=πkn5\theta_{n}=\frac{\pi k_{n}}{5} for all n2n \geq 2, where knk_{n} is an integer. θ1+2θ2=θ3\theta_{1}+2 \theta_{2}=\theta_{3}, so we may write θ1=πk15\theta_{1}=\frac{\pi k_{1}}{5} with k1k_{1} an integer. r3r4=r4r5r5=r42r3=r42r3\frac{r_{3}}{r_{4}}=\frac{r_{4}}{r_{5}} \Rightarrow r_{5}=\frac{r_{4}^{2}}{r_{3}}=r_{4}^{2} r_{3}, so r3=1.r3r4=2r4=12,r4=r32r2r2=12r_{3}=1 . \frac{r_{3}}{r_{4}}=2 \Rightarrow r_{4}=\frac{1}{2}, r_{4}=r_{3}^{2} r_{2} \Rightarrow r_{2}=\frac{1}{2}, and r3=r22r1r1=4r_{3}=r_{2}^{2} r_{1} \Rightarrow r_{1}=4. Therefore, the possible values of z1z_{1} are the nonreal roots of the equation x10410=0x^{10}-4^{10}=0, and the product of the eight possible values is 41042=48=65536\frac{4^{10}}{4^{2}}=4^{8}=65536. For these values of z1z_{1}, it is not difficult to construct a sequence which works, by choosing z2z_{2} nonreal so that z2=12\left|z_{2}\right|=\frac{1}{2}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.