Question

If $z_1,z_2,z_3$ are any three roots of the equation $Z^6={(z+1)}^6$, then $arg\left(\frac{z_1-z_3}{z_2+z_3}\right)$ can be equal to-

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{4}$
• D. $-\frac{\pi }{4}$

Answer 1

Answer: (B)

$\pi$

$z^6={(z+1)}^6;arg\left(\frac{z_1-z_3}{z_3+z_3}\right)$

By taking cube root on both sides

$\Rightarrow z^2={(z+1)}^2$

$z^2=z^2+2z+1$

$\therefore z=-\frac{1}{2}$

$\Rightarrow \left(\overline{z_1}\right)=-\frac{1}{2}[arg{z}_1+arg{z}_2+arg{z}_3]$

$arg{z}_1=-\frac{1}{2}[arg{z}_2+arg{z}_3]$

$\therefore arg\left(z_1\right)=\left(\overline{z_1}\right)=\frac{-\pi }{4}-\frac{1}{2}arg\left(\frac{z_3}{z_2+z_3}\right)$

$\Rightarrow arg\left(z_1\right)=\left(\overline{z_1}\right)=\frac{-\pi }{4}-\frac{1}{2}arg\left(\frac{z_3}{z_2+z_3}\right)$

$arg\left(z_1\right)+\frac{1}{2}arg\left(\frac{z_3}{z_2+z_3}\right)=\frac{\pi }{4}$

$arg\left(\frac{z_1-z_3}{z_2+z_3}\right)=\frac{\pi }{4}\times 4=\pi$

Hence, the answer is $\pi .$