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