Question

$\text{If}z=cos\frac{\pi }{4}+isin\frac{\pi }{6},\text{then}$

• A. $|z|=1,arg\left(z\right)=\frac{\pi }{4}$
• B. $|z|=1,arg\left(z\right)=\frac{\pi }{6}$
• C. $|z|=\frac{\sqrt{3}}{2},arg\left(z\right)=\frac{5\pi }{24}$
• D. $|z|=\frac{\sqrt{3}}{2},arg\left(z\right)=ta{n}^{-1}\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D

$|z|=\frac{\sqrt{3}}{2},arg\left(z\right)=ta{n}^{-1}\frac{1}{\sqrt{2}}$

$|z|=\frac{\sqrt{3}}{2},arg\left(z\right)=ta{n}^{-1}\frac{1}{\sqrt{2}}z=cos\frac{\pi }{4}+isin\frac{\pi }{6}\Rightarrow z=\frac{1}{\sqrt{2}}+\frac{1}{2}i\Rightarrow z=\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+\frac{1}{4}}\Rightarrow |z|=\sqrt{\frac{1}{2}+\frac{1}{4}}\Rightarrow |z|=\sqrt{\frac{3}{4}}\Rightarrow |z|=\frac{\sqrt{3}}{2}tan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=\frac{1}{\sqrt{2}}\Rightarrow \alpha =ta{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Since, the point z lies in the first quadrant.

Therefore, $arg\left(z\right)=\alpha =ta{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$