Complex Numbers Ques 98
Let a complex number $\alpha, \alpha \neq 1$, be a root of the equation
$ z^{p+q}-z^{p}-z^{q}+1=0 $
where, $p$ and $q$ are distinct primes. Show that either
$ \text { or } \quad \begin{aligned} & 1+\alpha+\alpha^{2}+\ldots+\alpha^{p-1}=0 \\ & 1+\alpha+\alpha^{2}+\ldots+\alpha^{q-1}=0 \end{aligned} $
but not both together.
(2002, 5M)
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Solution:
Formula:
Steps to determine nth roots of a complex number:
- Given, $\quad z^{p+q}-z^{p}-z^{q}+1=0$ $\quad$ …….(i)
$ \Rightarrow \quad\left(z^{p}-1\right)\left(z^{q}-1\right)=0 $
Since, $\alpha$ is root of Eq. (i), either $\alpha^{p}-1=0$ or $\alpha^{q}-1=0$
$\Rightarrow$ Either $\quad \frac{\alpha^{p}-1}{\alpha-1}=0$ or $\quad \frac{\alpha^{q}-1}{\alpha-1}=0 \quad[$ as $\alpha \neq 1]$
$\Rightarrow$ Either $\quad 1+\alpha+\alpha^{2}+\ldots+\alpha^{p-1}=0$
or $\quad 1+\alpha+\ldots+\alpha^{q-1}=0$
But $\quad \alpha^{p}-1=0$ and $\quad \alpha^{q}-1=0$ cannot occur simultaneously as $p$ and $q$ are distinct primes, so neither $p$ divides $q$ nor $q$ divides $p$, which is the requirement for $1=\alpha^{p}=\alpha^{q}$.
BETA
