Question

If $z_1 = 2 + 3i, z_2 = 3 - 2i, z_3 = - 1 - 2 i \sqrt{3}$ then which of the following hold good?

• A. $arg\left(\frac{z_{3}}{z_{2}}\right) = arg\left(\frac{z_{2}}{z_{1}}\right)$
• B. $arg\left(\frac{z_{3}}{z_{2}}\right) = 2arg\left(\frac{z_{3}- z_{1}}{z_{2}-z_{1}}\right)$
• C. $arg\left(\frac{z_{3}}{z_{2}}\right) = arg\left(\frac{z_{3}- z_{1}}{z_{2}-z_{1}}\right)$
• D. $arg\left(\frac{z_{3}}{z_{2}}\right) = \frac{1}{2}arg\left(\frac{z_{3}- z_{1}}{z_{2}-z_{1}}\right)$

Answer 1

Answer: (B)

$z_1 = 2 + 3i $
$\therefore |z_1| = \sqrt{13}$
$z_2 = 3 - 2i$
$\therefore |z_2| = \sqrt{13}$
and $z_3 = -1 - 2i \sqrt{3}$
$\therefore | z_3| = \sqrt{13}$
So $| z_1 | = | z_2 | = | z_3| = \sqrt{13}$
$\Rightarrow z_1, z_2, z_3$ lies on a circle of radius $\sqrt{13}$ & centre $(0, 0)$

Now, $arg\left(\frac{z_{3}}{z_{2}}\right) = 2 \,arg\left(\frac{z_{3}- z_{1}}{z_{3}-z_{1}}\right)$
$\therefore arg\left(\frac{z_{3}}{z_{2}}\right) = 2 \,arg \left(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\right)$