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

The correct option is B $\pi$

$z^6=(z+1{)}^6$
$\Rightarrow {\left(\frac{z+1}{z}\right)}^6=1=e^{i2k\pi }$
$\Rightarrow \frac{z+1}{z}=e^{ik\pi /3},k=0,1,2,3,4,5$
$\Rightarrow z=\frac{1}{e^{ik\pi /3}-1}$
$=\frac{1}{\cos \frac{k\pi }{3}+i\sin \frac{k\pi }{3}-1}$
$=\frac{1}{-2{\sin }^2\frac{k\pi }{6}+i2\sin \frac{k\pi }{6}\cdot \cos \frac{k\pi }{6}}$
$=\frac{1}{-2\sin \frac{k\pi }{6}(\sin \frac{k\pi }{6}-i\cos \frac{k\pi }{6})}$
$=\frac{\sin \frac{k\pi }{6}+i\cos \frac{k\pi }{6}}{-2\sin \frac{k\pi }{6}}$
$=-\frac{1}{2}(1+i\cot \frac{k\pi }{6})$
So, all the roots lie on the line $x=-\frac{1}{2}$ in Argand plane.
Thus, roots of $z^6=(z+1{)}^6$ are collinear.
$\Rightarrow \arg \left(\frac{z_1-z_3}{z_2-z_3}\right)=0$ or $\pi$