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Question
If\[z = \cos\frac{\pi}{4} + i \sin\frac{\pi}{6}\], then
Options
\[\left| z \right| = 1, \text { arg }(z) = \frac{\pi}{4}\]
\[\left| z \right| = 1, \text { arg }(z) = \frac{\pi}{6}\]
\[\left| z \right| = \frac{\sqrt{3}}{2},\text { arg }(z) = \frac{5\pi}{24}\]
\[\left| z \right| = \frac{\sqrt{3}}{2}, \text { arg }(z) = \tan^{- 1} \frac{1}{\sqrt{2}}\]
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Solution
\[z = \cos\frac{\pi}{4} + i\sin\frac{\pi}{6}\]
\[ \Rightarrow z = \frac{1}{\sqrt{2}} + \frac{1}{2}i\]
\[ \Rightarrow \left| z \right| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \frac{1}{4}}\]
\[ \Rightarrow \left| z \right| = \sqrt{\frac{1}{2} + \frac{1}{4}}\]
\[ \Rightarrow \left| z \right| = \sqrt{\frac{3}{4}}\]
\[ \Rightarrow \left| z \right| = \frac{\sqrt{3}}{2}\]
\[\tan \alpha = \left| \frac{\text { Im }(z)}{\text { Re }(z)} \right|\]
\[ = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \alpha = \tan^{- 1} \left( \frac{1}{\sqrt{2}} \right)\]
\[\text { Since, the point z lies in the first quadrant } . \]
\[\text { Therefore, } \arg(z) = \alpha = \tan^{- 1} \left( \frac{1}{\sqrt{2}} \right)\]
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