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प्रश्न
The amplitude of \[\frac{1 + i\sqrt{3}}{\sqrt{3} + i}\] is
पर्याय
\[\frac{\pi}{3}\]
\[- \frac{\pi}{3}\]
\[\frac{\pi}{6}\]
\[- \frac{\pi}{6}\]
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उत्तर
\[\frac{\pi}{6}\]
\[\text { Let }z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}\]
\[ \Rightarrow z=\frac{1 + i\sqrt{3}}{\sqrt{3} + i}\times\frac{\sqrt{3} - i}{\sqrt{3} - i}\]
\[ \Rightarrow z=\frac{\sqrt{3} + 2i - \sqrt{3} i^2}{3 - i^2}\]
\[ \Rightarrow z=\frac{\sqrt{3} + \sqrt{3} + 2i}{4}\]
\[ \Rightarrow z = \frac{2\sqrt{3} + 2i}{4}\]
\[ \Rightarrow z = \frac{\sqrt{3}}{2} + \frac{1}{2}i\]
\[\tan \alpha = \left| \frac{Im(z)}{Re(z)} \right|\]
\[ = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \alpha = \frac{\pi}{6}\]
\[\text { Since, z lies in the first quadrant } . \]
\[\text{Therefore,} arg(z)=\tan^{- 1}\left( \frac{1}{\sqrt{3}} \right)=\frac{\pi}{6}\]
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