Multiple choice

Find the value of $\displaystyle z^{14}+\dfrac{1}{z^{14}},$ where $z$ is a root of the equation $\dfrac{ z^2+1}{z}=1.$

  1. $1$
  2. $-1$
  3. $\omega$
  4. $\omega^2$
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A Correct answer
AI explanation

Rearranging the equation gives z squared minus z plus 1 equals 0, and multiplying by z plus 1 yields z cubed plus 1 equals 0, so z cubed equals minus 1. Since 14 leaves a remainder of 2 when divided by 3, z to the power of 14 equals z squared. Using the original equation, z squared plus 1 equals z, so z squared plus its reciprocal equals z divided by z squared plus 1 divided by z squared, which equals 1 divided by z plus 1 divided by z squared. Reversing the terms gives the reciprocal of z plus the reciprocal of z squared, which equals 1, so z to the power of 14 plus 1 divided by z to the power of 14 equals 1.

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