Question

Let $\omega$ be a complex number such that $2\omega +1=z$ where $z=\sqrt{-3}$, if $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=3k,$ then $k$ ie equal to.

• A. $-z$
• B. $z$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-z$

$2w+1=z$

$2w+1=\sqrt{3}i$

$w=\frac{-1+\sqrt{3}i}{2}$

$\Rightarrow w^2=\frac{-1-\sqrt{3}i}{2}=-w-1$

Now, $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=3k$ ...Given

$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& w& w^2\\ 1& w^2& w\end{array}\right|=3k$

$1(w^2-w^4)-1(w-w^2)+1(w^2-w)=3k$

$3w^2-3w=3k$

$3(w^2-w)=3k$

$k=w^2-w$

$k=-1-2w$

$=-1-(-1+\sqrt{3}i)$

$=-\sqrt{3}i$

$=-z$