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← JEE Main 2022 · 26 Jun · Shift 2
MathematicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 23

If z2+z+1=0z^2+z+1=0, z∈Cz\in\mathbb C, then ∣∑n=115(zn+(−1)n1zn)2∣\left|\displaystyle\sum_{n=1}^{15}\left(z^n+(-1)^n\frac1{z^n}\right)^2\right| is equal to ______.

Answer and worked solution

Official answer: 2

z=ωz=\omega (a cube root of unity), z−n=ω2nz^{-n}=\omega^{2n}.

Even nn. (ωn+ω2n)2(\omega^n+\omega^{2n})^2 is 11 if 3∤n3\nmid n and 44 if 3∣n3\mid n. Odd nn. (ωn−ω2n)2=ω2n+ωn−2(\omega^n-\omega^{2n})^2=\omega^{2n}+\omega^n-2 is −3-3 if 3∤n3\nmid n and 00 if 3∣n3\mid n.

Count for n=1n=1 to 15. 5(1)+2(4)+5(−3)+3(0)=−25(1)+2(4)+5(-3)+3(0)=-2; the modulus is 2.

Answer: 2