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प्रश्न
Express the following complex in the form r(cos θ + i sin θ):
tan α − i
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उत्तर
\[ \text { Let }z = \tan \alpha - i \]
\[ \because \tan \alpha\text { is periodic with period } \pi . \text { So, let us take } \]
\[\alpha \in [0, \frac{\pi}{2}) \cup ( \frac{\pi}{2}, \pi]\]
\[\text { Case I }: \]
\[z = \tan \alpha - i \]
\[ \Rightarrow \left| z \right| = \sqrt{\tan^2 + 1}\]
\[ = \left| \sec \alpha \right| \left[ \because 0 < \alpha < \frac{\pi}{2} \right]\]
\[ = \sec \alpha\]
\[\text { Let } \beta \text { be an acute angle given by }\tan \beta = \left| \frac{Im (z)}{Re(z)} \right|\]
\[\tan \beta = \frac{1}{\left| \tan \alpha \right|}\]
\[ = \left| \cot \alpha \right|\]
\[ = \cot \alpha\]
\[ = \tan \left( \frac{\pi}{2} - \alpha \right)\]
\[ \Rightarrow \beta = \frac{\pi}{2} - \alpha \]
\[\text { We can see that Re }(z) > 0 \text { and Im}(z) < 0 . \text { So, z lies in the fourth quadrant }. \]
\[ \therefore \arg(z) = - \beta = \alpha - \frac{\pi}{2}\]
\[\text { Thus, z in the polar form is given by }\]
\[z = \sec \alpha \left\{ \cos\left( \alpha - \frac{\pi}{2} \right) + i\sin \left( \alpha - \frac{\pi}{2} \right) \right\} \]
\[\text { Case II }: \]
\[z = \tan \alpha - i \]
\[ \Rightarrow \left| z \right| = \sqrt{\tan^2 + 1}\]
\[ = \left| \sec \alpha \right| \left[ \because \frac{\pi}{2} < \alpha < \pi \right]\]
\[ = - \sec \alpha\]
\[\text { Let } \beta \text { be an acute angle given by } \tan \beta = \left| \frac{Im (z)}{Re(z)} \right|\]
\[\tan \beta = \frac{1}{\left| \tan \alpha \right|}\]
\[ = \left| \cot \alpha \right|\]
\[ = - \cot \alpha\]
\[ = \tan \left( \alpha - \frac{\pi}{2} \right)\]
\[ \Rightarrow \beta = \alpha - \frac{\pi}{2}\]
\[\text{We can see that Re}(z) < 0 \text { and Im} (z) < 0 . So, z \text { lies in the third quadrant }. \]
\[ \therefore \arg(z) = \pi + \beta = \frac{\pi}{2} + \alpha\]
\[\text { Thus, z in the polar form is given by } \]
\[z = - \sec \alpha \left\{ \cos\left( \frac{\pi}{2} + \alpha \right) + i\sin \left( \frac{\pi}{2} + \alpha \right) \right\} \]
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