Question

If for the complex number $z$ satisfying $|z-2-2i|\le 1,$ the maximum value of $|3iz+6|$ is attained at $a+ib$, then $a+b$ is equal to

Answer 1

$|z-2-2i|\le 1$
$\Rightarrow z$ lies inside the circle with centre at $2+2i$ and radius $=1$, as shown in figure.
$|3iz+6|=|3i||z+\frac{6}{3i}|=3|z-2i|$
This is distance of $z$ from $2i$
Hence for maximum value $z=3+2i$ (Refer figure)
$\therefore a+b=5.$