can you please see if my solution is correct for the folowing equation

Question

The equation is

$ z^7+z^6+z^5+z^4+z^3+z^2+z=0 $

I tried to solve it that way $z(z^6+z^5+z^4+z^3+z^2+z+z+1)=0 $ then

No root was found algebraically so $z=0$ !!!! Am I right here? Special thanks to all of you.

Answer 1

Well you have this factored so you are on your way to solving it. If you have an expression in the form of $ab = 0$. Then we know that either $a = 0$, $b = 0$ or they are both $0$. You have: $z(z^6+z^5+z^4+z^3+z^2+z+z+1)=0$

So one root is $z = 0$. To find the other roots, you need to solve this equation: $z^6+z^5+z^4+z^3+z^2+z+z+1=0$.

I suspect that $0$ is the only real root.

Bob

Answer 2

HINT

From here

$$z(z^6+z^5+z^4+z^3+z^2+z+1)=0 \iff z=0 \,\lor\, z^6+z^5+z^4+z^3+z^2+z+1=0$$

then note that

$$z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^7-1}{z-1}$$

Answer 3

You also have six more roots in complex plane.

Note that $$ z^7-1=(z-1)(z^6+z^5+...+z+1)$$

Thus the seven-th roots of unity except $z=1$ are also solutions to your equation .