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प्रश्न
Express the following complex in the form r(cos θ + i sin θ):
1 − sin α + i cos α
उत्तर
\[\text { Let } z = \left( 1 - \sin\alpha \right) + i\cos\alpha . \]
\[ \because \text { sine and cosine functions are periodic functions with period } 2\pi . \]
\[\text { So, let us take } \alpha \in [0, 2\pi]\]
\[\text { Now, z } = 1 - \sin\alpha + i\cos\alpha\]
\[ \Rightarrow \left| z \right| = \sqrt{\left( 1 - \sin\alpha \right)^2 + \cos^2 \alpha} = \sqrt{2 - \sin\alpha} = \sqrt{2}\sqrt{1 - \sin\alpha}\]
\[ \Rightarrow \left| z \right| = \sqrt{2}\sqrt{\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)^2} = \sqrt{2}\left| \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right|\]
\[\text { Let } \beta \text { be an acute angle given by } \tan\beta = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|} . \text { Then }, \]
\[\tan\beta = \frac{\left| \cos\alpha \right|}{\left| 1 - \sin\alpha \right|} = \left| \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)^2} \right| = \left| \frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2} - \sin\frac{\alpha}{2}} \right|\]
\[ \Rightarrow \tan\beta = \left| \frac{1 + \tan\frac{\alpha}{2}}{1 - \tan\frac{\alpha}{2}} \right| = \left| \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right|\]
\[\text { Case I: When 0 } \leq \alpha < \frac{\pi}{2}\]
\[\text { In this case, we have }, \]
\[\cos\frac{\alpha}{2} > \sin\frac{\alpha}{2} \text { and } \frac{\pi}{4} + \frac{\alpha}{2} \in [\frac{\pi}{4}, \frac{\pi}{2})\]
\[ \Rightarrow \left| z \right| = \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\]
\[\text { and } \tan\beta = \left| \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right| = \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right)\]
\[ \Rightarrow \beta = \frac{\pi}{4} + \frac{\alpha}{2}\]
\[\text { Clearly, z lies in the first quadrant . Therefore }, \arg\left( z \right) = \frac{\pi}{4} + \frac{\alpha}{2}\]
\[\text { Hence, the polar form of z is } \]
\[\sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\left\{ \cos\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) + i\sin\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right\}\]
\[\text { Case II: When} \frac{\pi}{2} < \alpha < \frac{3\pi}{2}\]
\[\text { In this case, we have,}\]
\[\cos\frac{\alpha}{2} < \sin\frac{\alpha}{2} \text { and } \frac{\pi}{4} + \frac{\alpha}{2} \in \left( \frac{\pi}{2}, \pi \right)\]
\[ \Rightarrow \left| z \right| = \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right) = - \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\]
\[\text { and } \tan\beta = \left| \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right| = - \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) = \tan\left\{ \pi - \left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right\} = \tan\left( \frac{3\pi}{4} - \frac{\alpha}{2} \right)\]
\[ \Rightarrow \beta = \frac{3\pi}{4} - \frac{\alpha}{2}\]
\[\text { Clearly, z lies in the fourth quadrant . Therefore,} \arg\left( z \right) = - \beta = - \left( \frac{3\pi}{4} - \frac{\alpha}{2} \right) = \frac{\alpha}{2} - \frac{3\pi}{4}\]
\[\text { Hence, the polar form of z is} \]
\[ - \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\left\{ \cos\left( \frac{\alpha}{2} - \frac{3\pi}{4} \right) + i\sin\left( \frac{\alpha}{2} - \frac{3\pi}{4} \right) \right\}\]
\[\text { Case III: When } \frac{3\pi}{2} < \alpha < 2\pi\]
\[\text { In this case, we have, }\]
\[\cos\frac{\alpha}{2} < \sin\frac{\alpha}{2} and \frac{\pi}{4} + \frac{\alpha}{2} \in \left( \pi, \frac{5\pi}{4} \right)\]
\[ \Rightarrow \left| z \right| = \sqrt{2}\left| \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right| = - \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\]
\[\text { and } \tan\beta = \left| \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right| = \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) = - \tan\left\{ \pi - \left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right\} = \tan\left( \frac{\alpha}{2} - \frac{3\pi}{4} \right)\]
\[ \Rightarrow \beta = \frac{\alpha}{2} - \frac{3\pi}{4}\]
\[\text { Clearly, z lies in the first quadrant . Therefore,} \arg\left( z \right) = \beta = \frac{\alpha}{2} - \frac{3\pi}{4}\]
\[\text { Hence, the polar form of z is } \]
\[ - \sqrt{2}\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)\left\{ \cos\left( \frac{\alpha}{2} - \frac{3\pi}{4} \right) + i\sin\left( \frac{\alpha}{2} - \frac{3\pi}{4} \right) \right\}\]
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