Question

If ${}^{\prime }{\alpha }^{\prime }$ is a root of $x^4=1$ with negative principal argument, then the principal argument of $\Delta \left(\alpha \right)$ where
$\Delta \left(\alpha \right)=\left|\begin{array}{ccc}1& 1& 1\\ {\alpha }^n& {\alpha }^{n+1}& {\alpha }^{n+3}\\ \frac{1}{{\alpha }^{n+1}}& \frac{1}{{\alpha }^n}& 0\end{array}\right|$ is

• A. $\frac{5\pi }{14}$
  
• B. $-\frac{3\pi }{4}$
  
• C. $\frac{\pi }{4}$
  
• D. $-\frac{\pi }{4}$

Answer 1

The correct option is B $-\frac{3\pi }{4}$


Clearly $\alpha =-i$ where $i^2=-1$
So $\Delta \left(\alpha \right)={\alpha }^n\frac{1}{{\alpha }^n}\left|\begin{array}{ccc}1& 1& 1\\ 1& -i& i\\ i& 1& 0\end{array}\right|=1(-i)+1\left(i^2\right)+(1+i^2)=-1-i$
So, principal argument of $\Delta \left(\alpha \right)is-\frac{3\pi }{4}$