Question

If for $z$ as real or complex such that ${(1+z^2+z^4)}^8=C_0+C_1{z}^2+C_2{z}^4\cdot \cdot \cdot \cdot \cdot \cdot +C_{16}{z}^{32}$, then find $C_0+C_3+....+C_{15}+(C_2+C_5+C_8+C_{11}+C_{14})\omega +$ $(C_1+C_4+C_7+....+C_{16}){\omega }^2$, where $\omega$ is the cube root of unity .

• A. $0$
• B. $1$
• C. $\omega$
• D. none of the above

Answer 1

The correct option is A $0$

Substituting $z=\omega$
Hence
$(1+{\omega }^2+{\omega }^4)$
$=(1+{\omega }^2+{\omega }^3(\omega \left)\right)$
$=(1+{\omega }^2+\omega )$
$=0$
Hence $C_0+C_1{\omega }^2+C_2{\omega }^4...=0$
Rearranging the terms and replacing
${\omega }^{3n}=1$
${\omega }^4,{\omega }^{10},{\omega }^{16}..=\omega$ and
${\omega }^2,{\omega }^8,{\omega }^{14}={\omega }^2$
We get
$C_0+C_3+C_6+C_9+C_{12}+C_{15}+(C_2+C_5+C_8+C_{11}+C_{14})\omega +$

$(C_1+C_4+C_7+C_{10}+C_{13}+C_{16}){\omega }^2$
$=0$