Question

If\[z = \cos\frac{\pi}{4} + i \sin\frac{\pi}{6}\], then

विकल्प

• \[\left| z \right| = 1, \text { arg }(z) = \frac{\pi}{4}\]

• \[\left| z \right| = 1, \text { arg }(z) = \frac{\pi}{6}\]

• \[\left| z \right| = \frac{\sqrt{3}}{2},\text {  arg }(z) = \frac{5\pi}{24}\]

• \[\left| z \right| = \frac{\sqrt{3}}{2}, \text { arg }(z) = \tan^{- 1} \frac{1}{\sqrt{2}}\]

Answer 1

\[z = \cos\frac{\pi}{4} + i\sin\frac{\pi}{6}\]

\[ \Rightarrow z = \frac{1}{\sqrt{2}} + \frac{1}{2}i\]

\[ \Rightarrow \left| z \right| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \frac{1}{4}}\]

\[ \Rightarrow \left| z \right| = \sqrt{\frac{1}{2} + \frac{1}{4}}\]

\[ \Rightarrow \left| z \right| = \sqrt{\frac{3}{4}}\]

\[ \Rightarrow \left| z \right| = \frac{\sqrt{3}}{2}\]

\[\tan \alpha = \left| \frac{\text { Im }(z)}{\text { Re }(z)} \right|\]

\[ = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \alpha = \tan^{- 1} \left( \frac{1}{\sqrt{2}} \right)\]

\[\text { Since, the point z lies in the first quadrant } . \]

\[\text { Therefore, } \arg(z) = \alpha = \tan^{- 1} \left( \frac{1}{\sqrt{2}} \right)\]