Nonbrainiacs default....

$z_2 = -1 - i\sqrt{3}, z_3 = \sqrt{3} - i$ হলে দেখাও যে, $\text{arg}(z_2 z_3) = \text{arg} z_2 + \text{arg} z_3$

দেওয়া আছে, $z_2 = -1 - i\sqrt{3}, \quad z_3 = \sqrt{3} -i$
$\text{arg }z_2 = \text{arg }(-1 - i\sqrt{3})$
$= \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) - \pi$
$= \frac{\pi}{3} -\pi$
$= -2 \frac{\pi}{3}$
এবং $\text{arg } z_1 = \text{arg } (\sqrt{3} - i) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$
$z_2 z_3 = (-1 - i\sqrt{3})(\sqrt{3} - i) = -\sqrt{3} + i -3i + \sqrt{3} i^2 = -\sqrt{3} - 2i -\sqrt{3} = -2\sqrt{3} - 2i$
$\text{arg }(z_2 z_3) = \tan^{-1}\left(-\frac{2}{-2\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$
$\text{arg } z_2 + \text{arg } z_3 = -\frac{2\pi}{3} - \frac{\pi}{6} = -\frac{5\pi}{6} = \text{arg }(z_2 z_3) = \text{arg } z_2 + \text{arg } z_3$

$$\begin{array}{rcl} \text{H}_3\text{C} - \text{CH}_2 - \text{CH}_3 & \xrightarrow{\text{Cl}_2, \, \text{আলো}, \, 25^\circ\text{C}} & \underset{\text{স্ফুটনাঙ্ক}}{\text{H}_3\text{C} - \text{CH}_2 - \text{CH}_2 - \text{Cl}} + \text{H}_3\text{C} - \underset{\text{Cl}}{\text{CH}} - \text{CH}_3 \\ & & 45\% \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 55\% \\ & & (\text{স্ফুটনাঙ্ক } 47^\circ\text{C}) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (\text{স্ফুটনাঙ্ক } 36^\circ\text{C}) \\ \text{H}_3\text{C} - \text{CH}_2 - \text{CH}_3 & \xrightarrow{\text{Br}_2, \, \text{আলো}, \, 129^\circ\text{C}} & \text{H}_3\text{C} - \text{CH} - \text{CH}_2 - \text{Br} + \text{CH}_3 - \underset{\text{Br}}{\text{CH}} - \text{CH}_3 \\ & & 3\% \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 97\% \end{array}$$