Question

Let the complex number $z$ satisfies the inequalities ${\log }_{\frac{1}{2}}|z-3|+{\log }_2|z-3i|>0$, $|\mathrm{amp}(z+1-i)|\le \frac{\pi }{4}$ and $|z|\le 5$. If area of common region in which complex number $z$ lies is $\sqrt{\frac{a}{b}}\pi$ where $a,b$ are relatively prime numbers then find the value of $(a+b)$.

Answer 1

Answer: 164

$-{\log }_2|z-3|>{\log }_2|z-3|>0$
$\Rightarrow {\log }_2|z-3i|>{\log }_2|z-3|(z\ne 3,3i)$
$\text{ put }z=x+iy$
$|z-3i|>|z-3|$
$|x+i(y-3)|>|x-3+iy|$
$x^2+(y-3{)}^2>(x-3{)}^2+y^2$
$\Rightarrow x>y$
$|\mathrm{amp}(z-(-1+i))|\le \frac{\pi }{4}$
$\frac{-\pi }{4}\le \mathrm{amp}(z-(-1+i))\le \frac{\pi }{4}$
Region enclosed by z satisfying all the three given inequalities is sector $AOB$ in which point $(3,0)$ and the points on the line segment $OB$ are not included.
$\therefore$ Area of the sector $AOB=$ Area of quater circle $=\frac{25\pi }{4}=\sqrt{\frac{625}{16}}\pi$
$\therefore a+b=625+16=641$