Question

If $\left|\begin{array}{ccc}p& q& r\\ q& r& p\\ r& p& q\end{array}\right|=0$; where p, q, r are the moduli of non-zero complex numbers u, v, w respectively, find $arg\frac{w}{v}=$

• A. $arg{\left(\frac{v+u}{w+u}\right)}^2$
• B. $arg{\left(\frac{v-u}{w-u}\right)}^2$
• C. $arg{\left(\frac{w-u}{v-u}\right)}^2$
• D. $arg{\left(\frac{w+u}{v+u}\right)}^2$

Answer 1

Answer: (C)

$arg{\left(\frac{w-u}{v-u}\right)}^2$

$\left|\begin{array}{ccc}p& q& r\\ q& r& p\\ r& p& q\end{array}\right|=0$

$\Rightarrow p^3+q^3+r^3-3pqr=0$

$\Rightarrow (p+q+r)(p^2+q^2+r^2-pq-qr-rp)=0$

$p+q+r\ne 0$

$\frac{1}{2}[{(p-q)}^2+{(q-r)}^2+{(r-p)}^2]=0$

$\Rightarrow p=q=r$

$\Rightarrow v=u{e}^{\iota {\theta }_1}$

$w=u{e}^{\iota {\theta }_2}$

$arg\left(\frac{w}{v}\right)={\theta }_2-{\theta }_1$

$\frac{w}{v}=e^{\iota ({\theta }_1-{\theta }_2)}$

$\frac{w-u}{v-u}=\frac{u(1-e^{\iota {\theta }_2})}{u(1-e^{\iota {\theta }_1})}=\frac{2{\sin }^2\frac{{\theta }_2}{2}-\iota \left(2\sin \frac{{\theta }_2}{2}\cos \frac{{\theta }_2}{2}\right)}{2{\sin }^2\frac{{\theta }_1}{2}-\iota \left(2\sin \frac{{\theta }_1}{2}\cos \frac{{\theta }_1}{2}\right)}$

$=\frac{-\iota \left(2\sin \frac{{\theta }_2}{2}\right)}{-\iota 2\left(\sin \frac{{\theta }_1}{2}\right)}\frac{e^{\iota \frac{{\theta }_2}{2}}}{e^{\iota \frac{{\theta }_1}{2}}}$

$arg\left(\frac{w-u}{v-u}\right)=\frac{{\theta }_2-{\theta }_1}{2}$

$arg\frac{w}{v}=arg{\left(\frac{w-u}{v-u}\right)}^2$