Question

Let a complex number $\alpha (\alpha \ne 1)$ be a root of the equation $z^{12}-z^7-z^5+1=0$ and $S_n=\sum _{k=0}^{n-1}{\alpha }^k.$ Then which of the following is FALSE?

• A. $S_7=0$ and $S_5=0$
• B. $S_7=0$ but $S_5\ne 0$
• C. $S_5=0$ and $S_7\ne 0$
• D. $|\alpha |=1$

Answer 1

The correct option is A $S_7=0$ and $S_5=0$

Given $z^{12}-z^7-z^5+1=0$
$\Rightarrow z^7(z^5-1)-1(z^5-1)=0$
$\Rightarrow (z^7-1)(z^5-1)=0$
$\Rightarrow {\alpha }^7-1=0\text{or}{\alpha }^5-1=0$
($\because \alpha$ is a root)

$\text{Case : 1}$ when ${\alpha }^7-1=0$
$\Rightarrow (\alpha -1)({\alpha }^6+{\alpha }^5+{\alpha }^4+\cdots +1)=0)$
$\Rightarrow 1+\alpha +{\alpha }^2+\cdots +{\alpha }^6=0(\because \alpha \ne 1)$
$\Rightarrow S_7=0\cdots \left(1\right)$

$\text{Case : 2}$ when ${\alpha }^5-1=0$
$\Rightarrow (\alpha -1)({\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1)=0)$
$\Rightarrow 1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4=0(\because \alpha \ne 1)$
$\Rightarrow S_5=0\cdots \left(2\right)$

As $\alpha \ne 1,$ so ${\alpha }^7-1=0$ and ${\alpha }^5-1=0$ cannot occur simultaneously because $7$ and $5$ are distinct primes.
$\therefore$ Either $\left(1\right)$ is true or $\left(2\right)$ is true but not both.

Since, $\alpha$ is $n^{\text{th}}$ root of unity, therefore $|\alpha |=1$