Prove that $(1+z+z^2)(1+z+z^3)(1+z+z^4)=z(1+z)$

by user635988   Last Updated October 10, 2019 05:20 AM

We have $Z= \cos(2\pi/5)+i \sin(2\pi/5)$. Prove that:


I noticed that in each bracket contain $(1+z)$, so I let $t$ equal $(1+z)$ to simplify the equation into $(t+z^2)(t+z^3)(t+z^4)=zt$, but I don't know what to do next. I also tried to multiply the parentheses too but then the equation becomes crazier. I hope someone here could help me out. Thanks!

Answers 1

If you actually expand $(1+z+z^2)(1+z+z^3)(1+z+z^4) - z(1+z)$ out, you have:

$$z^9+z^8+2z^7+3z^6+4z^5+4z^4+4z^3+3z^2+2z+1$$ $$=5z^4+5z^3+5z^2+5z+5 \quad \text{(since $z^5=1$)}$$ $$=5 \left( \frac{z^5-1}{z-1} \right)$$ $$=0$$

Therefore $(1+z+z^2)(1+z+z^3)(1+z+z^4) = z(1+z)$.

Toby Mak
Toby Mak
October 10, 2019 05:17 AM

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