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

Question

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

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

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!

Answer 1

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

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

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

In addition, you can use your idea of making the substitution $t=1+z$, so that way you only have $8$ things to multiply.

Answer 2

Note that if $z=\cos(2\pi/5)+i\sin(2\pi/5)=e^{2\pi i/5}\not=1$ then $z^5=1$. Therefore, before taking the product, we write each factor in a convenient way: $$\begin{align} (1+z+z^2)&=\frac{1-z^3}{1-z}=\frac{z^2-1}{z^2(1-z)}=-\frac{1+z}{z^2},\\ (1+z+z^3)&=\frac{z^4(1+z+z^3)}{z^4}=\frac{1+z^2+z^4}{z^4}=\frac{1-z^6}{z^4(1-z^2)}=\frac{z(1-z)}{(1-z^2)}=\frac{z}{1+z},\\ (1+z+z^4)&=\frac{z(1+z+z^4)}{z}=\frac{1+z+z^2}{z}=-\frac{1+z}{z^3}=-z^2(1+z). \end{align}$$ Putting all together, it follows that $$(1+z+z^2)(1+z+z^3)(1+z+z^4)=-\frac{1+z}{z^2}\cdot \frac{z}{z+1}\cdot -z^2(1+z)=z(1+z).$$

Answer 3

Since you need to prove this holds for a specific value of $z$, just write it in exponential form and substitute.