Answer

  1. $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$

Solution:

$\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6}$

$\Rightarrow\text{z}=\frac{1}{\sqrt{2}}+\frac{1}{2}\text{i}$

$\Rightarrow|\text{z}|=\sqrt{\Big(\frac{1}{\sqrt{2}}\Big)^2+\frac{1}{4}}$

$\Rightarrow|\text{z}|=\sqrt{\frac{1}{2}+\frac{1}{4}}$

$\Rightarrow|\text{z}|=\sqrt{\frac{3}{4}}$

$\Rightarrow|\text{z}|=\frac{\sqrt{3}}{2}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\frac{1}{\sqrt{2}}$

$\Rightarrow\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$

Since, the point z lies in the first quadrant.

Therefore, $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$

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