JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Complex Numbers

The condition $|z-2-2i|\le 1$ describes a disk of radius $1$ centred at $2+2i$. Since $3iz+6=3i(z-2i)$, $|3iz+6|=3|z-2i|$, which is greatest when $z$ is farthest from the point $2i=(0,2)$. The centre $(2,2)$ lies at distance $2$ from $(0,2)$ along the direction $(1,0)$, so the farthest point of the disk is $(2,2)+(1,0)=(3,2)$, i.e. $z=3+2i$. Hence $a=3$, $b=2$ and $a+b=5$.