আমরা জানি, $\arg\left(\frac{z_2}{z_1}\right) = \arg(z_2) - \arg(z_1)$। এখানে, $$ \begin{aligned} \arg(z_2) &= \arg(\sqrt{3}+i) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \\ \arg(z_1) &= \arg(1-i) = \tan^{-1}(-1) = -\frac{\pi}{4} \end{aligned} $$ অতএব, $$ \begin{aligned} \arg\left(\frac{z_2}{z_1}\right) &= \frac{\pi}{6} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{6} + \frac{\pi}{4} = \frac{2\pi+3\pi}{12} = \frac{5\pi}{12} \end{aligned} $$