If z^2+z+1=0, z C, then _n=1^15|z^n+(-1)^n1z^n|^2 is equal to ____ Answer: 2 *(as recorded…

Question

If $z^2+z+1=0,\ z\in\mathbb C$, then $\displaystyle\sum_{n=1}^{15}\left|z^n+(-1)^n\frac1{z^n}\right|^2$ is equal to \_\_\_\_

Answer: 2 *(as recorded in the source --- see note below)*

Accepted Answer

$z=\omega$ or $\omega^2$ (a primitive cube root of unity). Computing the term for $n=1,\dots,6$ (the pattern repeats with period 6, since $z^n$ has period 3 and $(-1)^n$ has period 2):

- $n=1$: $z-z^{-1}=\omega-\omega^2=i\sqrt3$, squared modulus $=3$
- $n=2$: $z^2+z^{-2}=\omega^2+\omega=-1$, squared modulus $=1$
- $n=3$: $z^3-z^{-3}=1-1=0$, squared modulus $=0$
- $n=4$: $z^4+z^{-4}=z+z^{-1}=\omega+\omega^2=-1$, squared modulus $=1$
- $n=5$: $z^5-z^{-5}=z^2-z^{-2}=\omega^2-\omega=-i\sqrt3$, squared modulus $=3$
- $n=6$: $z^6+z^{-6}=1+1=2$, squared modulus $=4$

Sum over one period ($n=1$--$6$): $3+1+0+1+3+4=12$.

*(Note: carrying this pattern forward for $n=1$--$15$ (two full periods plus $n=13,14,15$, which repeat the $n=1,2,3$ values) gives $12+12+(3+1+0)=28$ by this direct computation, which does not match the source's recorded final answer of 2. The source provided no intermediate working for this item, so the discrepancy could not be resolved; both the computed value (28) and the source's stated answer (2) are noted here for transparency.)*

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