If a complex number z satisfies [2z+10+10i|≤ 5√3-5 then the least principal argument of z is-

Solution: (a); $|2 z+10+10 \mathrm{i}| \leq 5 \sqrt{3}-5$

$\begin{array}{l} \Rightarrow|z+5+5 i| \leq \frac{5 \sqrt{3}-5}{2} \Rightarrow|x+i y+5+5 i| \leq \frac{5 \sqrt{3}-5}{2} \\ \Rightarrow|(x+5)+i(y+5)| \leq \frac{5 \sqrt{3}-5}{2} \Rightarrow \sqrt{(x+5)^{2}+(y+5)^{2}} \leq \frac{5 \sqrt{3}-5}{2} \\ \therefore(x+5)^{2}+(y+5)^{2} \leq\left(\frac{5 \sqrt{3}-5}{2}\right)^{2} \ldots \ldots \ldots \ldots \text { (i) } \end{array}$

(i) indicates the inner side (including the perimeter) of a circle with center at $C(-5,-5)$ and radius, $r=\frac{5 \sqrt{3}-5}{2}$.

Here, $\mathrm{OC}=\sqrt{(0+5)^{2}+(0+5)^{2}}=5 \sqrt{2}$ units and $\mathrm{AC}=\mathrm{r}=\frac{5 \sqrt{3}-5}{2}$

Again, $\beta=\tan ^{-1}\left|\frac{-5}{-5}\right|=\frac{\pi}{4}$ and $\alpha=\sin ^{-1} \frac{A C}{O C}=\sin ^{-1}\left(\frac{\frac{5 \sqrt{3}-5}{2}}{5 \sqrt{2}}\right)=\frac{\pi}{12}$

Now, least principal argument, $\angle \mathrm{XOA}=-\left[\frac{\pi}{2}+\alpha+\beta\right]=-\left[\frac{\pi}{2}+\frac{\pi}{12}+\frac{\pi}{4}\right]=-\frac{5 \pi}{6}$ (Ans.)