Question

The common roots of the equations $z^3-(1+i)z^2+(1+i)z-i=0$ (where $i=\sqrt{-1})$ and $z^{2016}+z^{2017}-1=0$ are

• A. $-i,-\omega$
• B. $-\omega ,-{\omega }^2$
• C. $i,-\omega ,{\omega }^2$
• D. None of these

Answer 1

Answer: (D)

Given, $z^3-(1+i)z^2+(1+i)z-i=0$
$\to z^2(z-i)-z(z-i)+(z-i)=0$
$\Rightarrow (z-i)(z^2-z+1)=0$
$\Rightarrow (z-i)(z+\omega )(z+{\omega }^2)=0$
$\Rightarrow z=i,-\omega ,-{\omega }^2$
Now, putting these values in $z^{2016}+z^{2017}-1$ we have
(i) $(i{)}^{2016}+(i{)}^{2016}i-1=i^{4\lambda }+i{i}^{4\lambda }-1=i\ne 0$
$\therefore z=i$ is not a common root.
(ii) $(-\omega {)}^{2016}+(-\omega {)}^{2017}-1={\omega }^{3(672)}.\omega -1$
$=-\omega \ne 0$
$\therefore -\omega$ is not a common root.
(iii) $(-{\omega }^2{)}^{2016}+(-{\omega }^2{)}^{2017}-1$
$={\omega }^{4032}-{\omega }^{4034}-1$
$=1-{\omega }^2\ne 0$,
so the given equations have no common roots