Question

Let the complex number $z$ satisfies the inequalities $\log _{\frac{1}{2}}|z-3|+\log _2|z-3 i|>0$, $|\operatorname{amp}(z+1-i)| \leq \frac{\pi}{4}$ and $|z| \leq 5$. 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

$-\log _2| z -3|>\log _2| z -3|>0 $
$\Rightarrow \log _2| z -3 i|>\log _2| z -3|( z \neq 3,3 i) $
$\text { put } z = x +i y$
$| z -3 i|>| z -3|$
$| x +i( y -3)|>| x -3+i y | $
$x ^2+( y -3)^2>( x -3)^2+ y ^2 $
$\Rightarrow x > y$
$|\operatorname{amp}( z -(-1+i))| \leq \frac{\pi}{4}$
$\frac{-\pi}{4} \leq \operatorname{amp}( z -(-1+i)) \leq \frac{\pi}{4}$
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 $=\frac{25 \pi}{4}=\sqrt{\frac{625}{16}} \pi$
$\therefore a+b=625+16=641$