Question

S₁ :    x² + y² < 16

S₂ :    $\mathrm{lm}\left(\frac{(\mathrm{z}-1+\sqrt{3}\mathrm{i})(1+\sqrt{3}\mathrm{i})}{4}\right)>03$

or       $\mathrm{lm}\left(\frac{\{\mathrm{z}-(1-\sqrt{3}\mathrm{i}\left)\right\}\{1+\sqrt{3}\mathrm{i}\}}{4}\right)>0$

or       $\mathrm{lm}\left(\frac{(1+\sqrt{3}\mathrm{i})\mathrm{z}-4}{4}\right)>0\Rightarrow \sqrt{3}\mathrm{x}+\mathrm{y}>0$

S₃ :       x > 0                             

(i)         |1 – 3i – z| = |z – 1 + 3i| = |z – (1 – 3i)|

            Distance between [z and (1, –3)] = $\left|\frac{\sqrt{3}-3}{\sqrt{3+1}}\right|=\frac{3-\sqrt{3}}{\sqrt{3+1}}$    

(ii)         Area of S = $\frac{1}{2}\times 16\times \frac{5\mathrm{\pi }}{6}=\frac{20\mathrm{\pi }}{3}$  sq. unit

S₁ :    x² + y² < 16

S₂ :     $\mathrm{lm}\left(\frac{(\mathrm{z}-1+\sqrt{3}\mathrm{i})(1+\sqrt{3}\mathrm{i})}{4}\right)>03$

or        $\mathrm{lm}\left(\frac{\{\mathrm{z}-(1-\sqrt{3}\mathrm{i}\left)\right\}\{1+\sqrt{3}\mathrm{i}\}}{4}\right)>0$

or        $\mathrm{lm}\left(\frac{(1+\sqrt{3}\mathrm{i})\mathrm{z}-4}{4}\right)>0\Rightarrow \sqrt{3}\mathrm{x}+\mathrm{y}>0$

S₃ :       x > 0                             

(i)         |1 – 3i – z| = |z – 1 + 3i| = |z – (1 – 3i)|

            Distance between [z and (1, –3)] = $\left|\frac{\sqrt{3}-3}{\sqrt{3+1}}\right|=\frac{3-\sqrt{3}}{\sqrt{3+1}}$        

(ii)         Area of S = $\frac{1}{2}\times 16\times \frac{5\mathrm{\pi }}{6}=\frac{20\mathrm{\pi }}{3}$ sq. unit